Category: ML Aggarwal Solutions

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ML Aggarwal Class 12 Maths Solutions Section B Chapter 1 Vectors Ex 1.3

Very short answer type questions (1 to 13) :

Question 1.
(i) If \(\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}\) and \(\vec{b}=3 \hat{i}+5 \hat{j}-2 \hat{k}\), then find \(|\vec{a} \times \vec{b}|\). (NCERT)
(ii) Find the magnitude of \(\vec{a}\), where \(\vec{a}=(\hat{i}+3 \hat{j}-2 \hat{k}) \times(-\hat{i}+3 \hat{k})\).
(iii) If \(\vec{a}=3 \hat{i}-5 \hat{j}\), \(\vec{b}=6 \hat{i}+3 \hat{j}\) are two vectors and \(\vec{c}\) is a vector such that \(\vec{c}=\vec{a} \times \vec{b}\), then find \(|\vec{a}|:|\vec{b}|:|\vec{c}|\).
Solution:
(i) ∴ \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 1 & 3 \\
3 & 5 & -2
\end{array}\right|\)
= \(\hat{i}(-2-15)-\hat{j}(-4-9)+\hat{k}(10-3)\)
= \(-17 \hat{i}+13 \hat{j}+7 \hat{k}\)
Thus \(|\vec{a} \times \vec{b}|=\sqrt{(-17)^2+(13)^2+7^2}\)
= \(=\sqrt{289+169+49}=\sqrt{507}\)

(ii) \(\vec{a}=(\hat{i}+3 \hat{j}-2 \hat{k}) \times(-\hat{i}+3 \hat{k})\)
∴ \(\vec{a}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 3 & -2 \\
-1 & 0 & 3
\end{array}\right|\)
= \(\hat{i}(9-0)-\hat{j}(3-2)+\hat{k}(0+3)\)
= \(9 \hat{i}-\hat{j}+3 \hat{k}\)
∴ \(|\vec{a}|=\sqrt{9^2+(-1)^2+3^2}\)
= \(\sqrt{81+1+9}=\sqrt{91}\)

(iii) Given \(\vec{a}=3 \hat{i}-5 \hat{j}\) ;
\(\vec{b}=6 \hat{i}+3 \hat{j}\)
∴ \(\vec{c}=\vec{a} \times \vec{b}\)
= \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -5 & 0 \\
6 & 3 & 0
\end{array}\right|\)
= \(\hat{i}(0-0)-\hat{j}(0-0)+\hat{k}(9+30)\)
= 39 \(\hat{k}\)
Thus \(|\vec{c}|=|39 \hat{k}|\)
= 39 × 1 = 39
\(|\vec{a}|=\sqrt{3^2+(-5)^2}\)
= \(\sqrt{9+25}=\sqrt{34}\)
\(|\vec{b}|=\sqrt{6^2+3^2}\)
= \(\sqrt{36+9}=\sqrt{45}\)
Thus, \(|\vec{a}|:|\vec{b}|:|\vec{c}|=\sqrt{34}: \sqrt{45}: 39\).

Question 1 (old).
(ii) If \(\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}\) and \(\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}\), then find \(|\vec{a} \times \vec{b}|\).
Solution:
Given \(\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}\)
and \(\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}\)
∴ \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -7 & 7 \\
3 & -2 & 2
\end{array}\right|\)
= \(\hat{i}(0)-\hat{j}(2-21)+\hat{k}(-2+21)\)
= \(19 \hat{j}+19 \hat{k}\)
∴ \(|\vec{a} \times \vec{b}|=\sqrt{(19)^2+(19)^2}=19 \sqrt{2}\)

Question 2.
(i) Find the angle between two vectors \(\vec{a} \text { and } \vec{b}\) with magnitudes 1 and 2 respectively when \(|\vec{a} \times \vec{b}|=\sqrt{3}\).
(ii) If vectors \(\vec{a} \text { and } \vec{b}\) are such that \(|\vec{a}|=3,|\vec{b}|=\frac{2}{3}\) and \(\vec{a} \times \vec{b}\) is a unit vector then find the angle between \(\vec{a} \text { and } \vec{b}\).
(iii) If \(|\vec{a}|\) = 2, \(|\vec{b}|\) = 7 and \(\vec{a} \times \vec{b}=3 \hat{i}+2 \hat{j}+6 \hat{k}\), find the angle between \(\vec{a} \text { and } \vec{b}\).
Solution:
(i) Given \(|\vec{a}|\) = 1 ;
\(|\vec{b}|\) = 2
also, \(\sqrt{3}=|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}|\) sin θ
[∵ \(|\hat{\eta}|\) = 1]
where θ be the angle between \(\vec{a} \text { and } \vec{b}\)
⇒ √3 = 1 × 2 × sin θ
⇒ sin θ = \(\frac{\sqrt{3}}{2}\)
⇒ θ = \(\frac{\pi}{3}\)

(ii) Given \(|\vec{a}|\) =3 ;
\(|\vec{b}|\) = \(\frac{2}{3}\)
and \(\vec{a} \times \vec{b}\) is a unit vector
or \(\vec{a} \times \vec{b}\) = 1
⇒ \(|\vec{a}||\vec{b}|\) sin θ = 1,
where θ be the angle between \(\vec{a} \text { and } \vec{b}\)
⇒ 3 × \(\frac{2}{3}\) sin θ = 1
sin θ = \(\frac{1}{2}\)
θ = \(\frac{\pi}{6}\)

(iii) Given \(|\vec{a}|\) = 2 ;
\(|\vec{b}|\) = 7
and \(\vec{a} \times \vec{b}=3 \hat{i}+2 \hat{j}+6 \hat{k}\)
= \(\sqrt{9+4+36}\)
= \(\sqrt{49}\) = 7
Let θ be the angle between \(\vec{a} \text { and } \vec{b}\)
Then sin θ = \(\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}\)
∴ sin θ = \(\frac{7}{2 \times 7}=\frac{1}{2}\)
θ = 30°, 150°

Question 3.
Find the angle between two vectors \(\vec{a} \text { and } \vec{b}\) if \(|\vec{a} \times \vec{b}|=\vec{a} \cdot \vec{b}\). (NCERT)
Solution:
Let θ be the angle between \(\vec{a} \text { and } \vec{b}\)
Then \(|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}|\) sin θ
[∵ \(|\hat{\eta}|\) = 1]
and \(|\vec{a} \cdot \vec{b}|=|\vec{a}||\vec{b}|\) cos θ
Since \(|\vec{a} \times \vec{b}|=|\vec{a} \cdot \vec{b}|\) (given)
⇒ \(|\vec{a}||\vec{b}|\) sin θ = \(|\vec{a}||\vec{b}|\) cos θ
⇒ tan θ = 1
⇒ θ = \(\frac{\pi}{4}\)

Question 4.
(i) Find \(\vec{a} \cdot \vec{b}\) if \(|\vec{a}|\) = 2, \(|\vec{b}|\) = 5 and \(|\vec{a} \times \vec{b}|\) = 8.
(ii) If \(|\vec{a}|\) = 4 ; \(|\vec{b}|\) = 2 and angle between \(|\vec{a}|\) and \(|\vec{b}|\) is \(\frac{\pi}{3}\), find \((\vec{a} \times \vec{b})^2\).
(iii) If vectors \(\vec{a} \text { and } \vec{b}\) are such that \(|\vec{a}|\) = \(\frac{1}{2}\), \(|\vec{b}|=\frac{4}{\sqrt{3}}\) and \(|\vec{a} \times \vec{b}|=\frac{1}{\sqrt{3}}\) then find \(|\vec{a} \cdot \vec{b}|\).
Solution:
(i) Given, \(|\vec{a}|\) = 2,
\(|\vec{b}|\) = 5
and \(|\vec{a} \times \vec{b}|\) = 8
By Lagranges identity, we have
\(|\vec{a} \times \vec{b}|^2=|\vec{a}||\vec{b}|^2-(\vec{a} \cdot \vec{b})^2\)
⇒ 64 = 4 × 25 – \((\vec{a} \cdot \vec{b})^2\)
⇒ \((\vec{a} \cdot \vec{b})^2\) = 36
⇒ \(\vec{a} \cdot \vec{b}\) = 6 (Taking +ve sign)

(ii) Given \(|\vec{a}|\) = 4 ;
\(|\vec{b}|\) = 2
since angle between \(|\vec{a}|\) and \(|\vec{b}|\) be \(\frac{\pi}{3}\)

(iii) Given \(|\vec{a}|\) = \(\frac{1}{2}\) ;
\(|\vec{b}|=\frac{4}{\sqrt{3}}\)
and \(|\vec{a} \times \vec{b}|=\frac{1}{\sqrt{3}}\)

Question 4 (old).
(ii) If \(|\vec{a}|\) = 5, \(|\vec{b}|\) = 13 and \(|\vec{a} \times \vec{b}|\) = 25, then find \(\vec{a} \cdot \vec{b}\).
Solution:
Given \(|\vec{a}|\) = 5,
\(|\vec{b}|\) = 13
and \(|\vec{a} \times \vec{b}|\) = 25
If θ be the angle between \(\vec{a} \text { and } \vec{b}\)
sin θ = \(\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}\)
= \(\frac{25}{5 \times 13}=\frac{5}{13}\)
∴ cos θ = \(\sqrt{1-\frac{25}{169}}=\frac{12}{13}\)
∴ \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}|\) cos θ
= 5 × 13 × \(\frac{12}{13}\) = 60.

Question 5.
(i) Find the value of λ if \((2 \hat{i}+6 \hat{j}+14 \hat{k}) \times(\hat{i}-\lambda \hat{j}+7 \hat{k})=\overrightarrow{0}\).
(ii) Find the value of p if \((2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+3 \hat{j}+p \hat{k})=\overrightarrow{0}\).
(iii) Find λ and µ \((2 \hat{i}+6 \hat{j}+27 \hat{k}) \times(\hat{i}+\lambda \hat{j}+\mu \hat{k})=\overrightarrow{0}\). (NCERT)
Solution:
(i) Let \(\vec{a}=2 \hat{i}+6 \hat{j}+14 \hat{k}\) ;
\(\vec{b}=(\hat{i}-\lambda \hat{j}+7 \hat{k})\)
∴ \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 6 & 14 \\
1 & -\lambda & 7
\end{array}\right|\)
= \(\hat{i}(42+14 \lambda)-\hat{j}(14-14)+\hat{k}(-2 \lambda-6)\)
= \(\hat{i}(42+14 \lambda)+0 \hat{j}+(-2 \lambda-6) \hat{k}\)
It is given that \(\vec{a} \times \vec{b}=\overrightarrow{0}\)
⇒ \((42+14 \lambda) \hat{i}+0 \hat{j}+(-2 \lambda-6) \hat{k}\) = \(0 \hat{i}+0 \hat{j}+0 \hat{k}\)
∴ 42 + 14λ = 0
⇒ λ = – 3
and – 2λ – 6 = 0
λ = 3

(ii) Let \(\vec{a}=2 \hat{i}+6 \hat{j}+27 \hat{k}\) ;
\(\vec{b}=\hat{i}+3 \hat{j}+p \hat{k}\)

(iii) Let \(\vec{a}=2 \hat{i}+6 \hat{j}+27 \hat{k}\)
and \(\vec{b}=\hat{i}+\lambda \hat{j}+\mu \hat{k}\)
∴ \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 6 & 27 \\
1 & \lambda & \mu
\end{array}\right|\)
= \(\hat{i}(6 \mu-27 \lambda)-\hat{j}(2 \mu-27)+\hat{k}(2 \lambda-6)\)
also \(\vec{a} \times \vec{b}=\overrightarrow{0}\)
∴ \((6 \mu-27 \lambda) \hat{i}-\hat{j}(2 \mu-27)+\hat{k}(2 \lambda-6)=\overrightarrow{0}=0 \hat{i}+0 \hat{j}+0 \hat{k}\) ……………….(1)
Thus 6µ – 27λ = 0 ………………………(1)
– (2µ – 27) = 0
⇒ µ = \(\frac{27}{2}\)
and 2λ – 6 = 0
⇒ λ = 3
Also, λ = 3, µ = \(\frac{27}{2}\) satisfies eqn. (1).

Question 6.
(i) Find a unit vector perpendicular to the two vectors \(\hat{i}+2 \hat{j}-\hat{k}\) and \(2 \hat{i}+3 \hat{j}+\hat{k}\). (ISC 2004)
(ii) Find a unit vector perpendicular to \(\vec{a} \text { and } \vec{b}\), where \(\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}\) and \(\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}\).
Solution:
(i) Let \(\vec{a}=\hat{i}+2 \hat{j}-\hat{k}\) ;
\(\vec{b}=2 \hat{i}+3 \hat{j}+\hat{k}\)

(ii) Given \(\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}\)
and \(\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}\)
∴ \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -7 & 7 \\
3 & -2 & 2
\end{array}\right|\)
= \(\hat{i}(-14+14)-\hat{j}(2-21)+\hat{k}(-2+21)\)
= \(0 \hat{i}+19 \hat{j}+19 \hat{k}\)
Thus unit vector = \(\pm \frac{\vec{a} \times \vec{b}}{|\vec{a}+\vec{b}|}\)
= \(\pm \frac{19(\hat{j}+\hat{k})}{\sqrt{361+361}}\)
= \(\pm \frac{(\hat{j}+\hat{k})}{\sqrt{2}}\)

(ii) Find a unit vector perpendicular to the plane of \(\vec{a} \text { and } \vec{b}\), where \(\vec{a}=3 \hat{i}+2 \hat{j}+5 \hat{k}\) and \(\vec{b}=\hat{i}-3 \hat{j}+\hat{k}\).
Solution:
Given latex]\vec{a}=3 \hat{i}+2 \hat{j}+5 \hat{k}[/latex]
and \(\vec{b}=\hat{i}-3 \hat{j}+\hat{k}\)

Question 7.
(i) What conclusion can you draw about vectors \(\vec{a} \text { and } \vec{b}\) when \(\vec{a} \times \vec{b}=\overrightarrow{0}\) and \(\vec{a} \cdot \vec{b}\) = 0? (NCERT)
(ii) If either \(\vec{a}\) = 0 or \(\vec{b}\) = 0, then \(\vec{a} \times \vec{b}=\overrightarrow{0}\). Is the converse true? Justify your answer with an example. (NCERT)
Solution:
(i) When \(\vec{a} \times \vec{b}=\overrightarrow{0}\)
⇒ \(\vec{a}\) = 0
or \(\vec{a} \| \vec{b}\)
or \(\vec{a} \cdot \vec{b}\) = 0
⇒ \(\vec{a}=\overrightarrow{0}\)
or \(\vec{b}=\overrightarrow{0}\)
or \(\vec{a} \perp \vec{b}\)
Now a vector can never be parallel and perpendicular simultaneously.
Thus \(\vec{a}=\overrightarrow{0}\) or \(\vec{b}=\overrightarrow{0}\) = 0.

(ii) Given \(\vec{a}=\overrightarrow{0}\)
or \(\vec{b}=\overrightarrow{0}\)
∴ \(\vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \hat{n}\) = 0
Let \(\vec{a}=\hat{i}\) ;
b = 2 \(\hat{i}\)
Here \(\)
but neither \(\vec{a}=\overrightarrow{0}\) nor \(\vec{b}=\overrightarrow{0}\).

Question 7 (old).
(i) show that \(\vec{a} \times \vec{b}=\vec{a} \times \vec{c}\) may not imply that \(\vec{b}=\vec{c}\).
(ii) If \(\vec{b}=\vec{c}\), is \(\vec{a} \times \vec{b}=\vec{c} \times \vec{a}\) ?
Solution:
(i)

(ii) Not necessary ;

Question 8.
Find λ such that \(\vec{a}=\hat{i}+\lambda \hat{j}+3 \hat{k}\) and \(\vec{b}=3 \hat{i}+2 \hat{j}+9 \hat{k}\) are parallel.
Solution:
Given \(\vec{a}=\hat{i}+\lambda \hat{j}+3 \hat{k}\)
and \(\vec{b}=3 \hat{i}+2 \hat{j}+9 \hat{k}\)
Since \(\vec{a} \text { and } \vec{b}\) are parallel.
∴ \(\vec{a} \times \vec{b}=\overrightarrow{0}\) …………….(1)
Here, \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & \lambda & 3 \\
3 & 2 & 9
\end{array}\right|\)
= \(\hat{i}(9 \lambda-6)-\hat{j}(9-9)+\hat{k}(2-3 \lambda)\)
= \((9 \lambda-6) \hat{i}+(2-3 \lambda) \hat{k}\)
∴ from (1) ; we have
\((9 \lambda-6) \hat{i}+(2-3 \lambda) \hat{k}\)
= \(\overrightarrow{0}=0 \hat{i}+0 \hat{j}+0 \hat{k}\)
∴ 9λ – 6 = 0
⇒ λ = \(\frac{2}{3}\)
and 2 – 3λ = 0
⇒ λ = \(\frac{2}{3}\)

Question 9.
(i) \((\hat{i} \times \hat{j}) \cdot \hat{k}+\hat{i} \cdot \hat{j}\)
(ii) \((\hat{j} \times \hat{k}) \cdot \hat{i}+(\hat{i} \times \hat{k}) \cdot \hat{j}+(\hat{i} \times \hat{j}) \cdot \hat{k}\)
Solution:
Since \(\hat{i}, \hat{j}, \hat{k}\) forms right handed orthogonal system
∴ \(\hat{i} \times \hat{j}=\hat{k}\)
\(\hat{j} \times \hat{k}=\hat{i}\)
\(\hat{k} \times \hat{i}=\hat{j}\)
and \(\hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}\) = 0

(i) \((\hat{i} \times \hat{j}) \cdot \hat{k}+\hat{i} \cdot \hat{j}\)
\(\hat{k} \cdot \hat{k}+\hat{i} \cdot \hat{j}\)
= 1 + 0 = 1
[∵ \(\hat{i} \cdot \hat{i}=\hat{j} \cdot \hat{j}=\hat{k} \cdot \hat{k}\) = 1
and \(\hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{k}=\hat{k} \cdot \hat{i}\) = 0]

(ii) \((\hat{j} \times \hat{k}) \cdot \hat{i}+(\hat{i} \times \hat{k}) \cdot \hat{j}+(\hat{i} \times \hat{j}) \cdot \hat{k}\)
\(\hat{i} \cdot \hat{i}+(-\hat{j}) \cdot \hat{j}+(\hat{k} \cdot \hat{k})\)
= 1 – 1 + 1 = 0

Question 10.
Find the area of the parallelogram whose adjacent sides are represented by the vectors
(i) \(3 \hat{i}+\hat{j}+4 \hat{k} \text { and } \hat{i}-\hat{j}+\hat{k}\) (NCERT)
(ii) \(2 \hat{i}-3 \hat{k} \text { and } 4 \hat{j}+2 \hat{k}\)
Solution:
(i) Let \(\vec{a}=3 \hat{i}+\hat{j}+4 \hat{k}\)
and \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\)

(ii) Let \(\vec{a}=2 \hat{i}-3 \hat{k}\)
and \(\vec{b}=4 \hat{j}+2 \hat{k}\)

Question 11.
Find the area of a triangle whose two sides are represented by the vectors \(3 \hat{i}+\hat{j}+4 \hat{k}\) and \(\hat{i}-\hat{j}+\hat{k}\).
Solution:
Let \(\vec{a}=3 \hat{i}+\hat{j}+4 \hat{k}\)
and \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\)
∴ \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & 1 & 4 \\
1 & -1 & 1
\end{array}\right|\)
= \(\hat{i}(1+4)-\hat{j}(3-4)+\hat{k}(-3-1)\)

∴ \(\vec{a} \times \vec{b}=5 \hat{i}+\hat{j}-4 \hat{k}\)
Thus, \(|\vec{a} \times \vec{b}|=\sqrt{5^2+1^2+(-4)^2}\)
= \(\sqrt{25+1+16}=\sqrt{42}\)

Therefore, area of triangle = \(\frac{1}{2}|\vec{a} \times \vec{b}|\)
= \(\frac{1}{2} \sqrt{42}\) sq. units

Question 12.
(i) If \(\vec{a}=4 \hat{i}+3 \hat{j}+2 \hat{k}\) and \(\vec{b}=3 \hat{i}+2 \hat{k}\), then find \(|\vec{b} \times 2 \vec{a}|\).
(ii) If \(\vec{a}=\hat{i}+\hat{j}-3 \hat{k}\) and \(\vec{b}=\hat{j}+2 \hat{k}\), then find \(|2 \vec{b} \times \vec{a}|\).
Solution:
(i) Given, \(\vec{a}=4 \hat{i}+3 \hat{j}+2 \hat{k}\)
and \(\vec{b}=3 \hat{i}+2 \hat{k}\)

(ii) Given, \(\vec{a}=\hat{i}+\hat{j}-3 \hat{k}\) ;
\(\vec{b}=\hat{j}+2 \hat{k}\)

∴ \(2 \vec{b}=2 \hat{j}+4 \hat{k}\)
\(2 \vec{b} \times \vec{a}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
0 & 2 & 4 \\
1 & 1 & -3
\end{array}\right|\)
= \(\hat{i}(-6-4)-\hat{j}(-4)+\hat{k}(-2)\)
= \(-10 \hat{i}+4 \hat{j}-2 \hat{k}\)

∴ \(|2 \vec{b} \times \vec{a}|=\sqrt{(-10)^2+(4)^2+(-2)^2}\)
= \(\sqrt{100+16+4}\)
= \(\sqrt{120}=2 \sqrt{30}\)

Question 13.
(i) If \(\vec{a}=3 \hat{i}-\hat{j}-2 \hat{k}\) and \(\vec{b}=2 \hat{i}+3 \hat{j}+\hat{k}\), then find \((\vec{a}+2 \vec{b}) \times(2 \vec{a}-\vec{b})\).
(ii) Taking \(\vec{a}=2 \hat{i}-3 \hat{j}-\hat{k}\) and \(\vec{b}=\hat{i}+4 \hat{j}-2 \hat{k}\) verify that \(\vec{a} \times \vec{b} \neq \vec{b} \times \vec{a}\).
Solution:
(i) Given \(\vec{a}=3 \hat{i}-\hat{j}-2 \hat{k}\) ;
\(\vec{b}=2 \hat{i}+3 \hat{j}+\hat{k}\)
\(\vec{a}+2 \vec{b}=(3 \hat{i}-\hat{j}-2 \hat{k})+2(2 \hat{i}+3 \hat{j}+\hat{k})\)
= \(7 \hat{i}+5 \hat{j}+0 \hat{k}\)

\(2 \vec{a}-\vec{b}=2(3 \hat{i}-\hat{j}-2 \hat{k})-(2 \hat{i}+3 \hat{j}+\hat{k})\)
\(4 \hat{i}-5 \hat{j}-5 \hat{k}\)

Thus \((\vec{a}+2 \vec{b}) \times(2 \vec{a}-\vec{b})\)
= \(\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
7 & 5 & 0 \\
4 & -5 & -5
\end{array}\right|\)
= \(\hat{i}(-25-0)-\hat{j}(-35-0)+\hat{k}(-35-20)\)
= \(-25 \hat{i}+35 \hat{j}-55 \hat{k}\)

(ii) Given \(\vec{a}=2 \hat{i}-3 \hat{j}-\hat{k}\)
and \(\vec{b}=\hat{i}+4 \hat{j}-2 \hat{k}\)

∴ \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & -3 & -1 \\
1 & 4 & -2
\end{array}\right|\)
= \(\hat{i}(6+4)-\hat{j}(-4+1)+\hat{k}(8+3)\)
= \(10 \hat{i}+3 \hat{j}+11 \hat{k}\)

∴ \(\vec{b} \times \vec{a}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & 4 & -2 \\
2 & -3 & -1
\end{array}\right|\)
= \(\hat{i}(-4-6)-\hat{j}(-1+4)+\hat{k}(-3-8)\)
= \(-10 \hat{i}-3 \hat{j}-11 \hat{k}\)

∴ \(\vec{a} \times \vec{b} \neq \vec{b} \times \vec{a}\)

Question 14.
If θ is the angle between two vectors \(\hat{i}-2 \hat{j}+3 \hat{k}\) and \(3 \hat{i}-2 \hat{j}+\hat{k}\), find sin θ.
Solution:
Let \(\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}\)
and \(\vec{b}=3 \hat{i}-2 \hat{j}+\hat{k}\)

Question 15.
If \(\vec{a}=3 \hat{i}+\hat{j}+2 \hat{k}\) and \(\vec{b}=2 \hat{i}-2 \hat{j}+4 \hat{k}\), then find :
(i) the sine of the angle between \(\vec{a} \text { and } \vec{b}\). (NCERT Exemplar)
(ii) a unit vector perpendicular to both the vectors \(\vec{a} \text { and } \vec{b}\).
Solution:
(i) Given \(\vec{a}=3 \hat{i}+\hat{j}+2 \hat{k}\)
and \(\vec{b}=2 \hat{i}-2 \hat{j}+4 \hat{k}\)
Let θ be the angle between \(\vec{a} \text { and } \vec{b}\)

(ii) Here,
\(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & 1 & 2 \\
2 & -2 & 4
\end{array}\right|\)
= \(8 i-8 j-8 k\)
\(|\vec{a} \times \vec{b}|=\sqrt{8^2+(-8)^2+(-8)^2}\)
= 8√3
Thus required unit vector ⊥to both \(\vec{a} \text { and } \vec{b}\)
= \(\pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} \pm \frac{8(\hat{i}-\hat{j}-\hat{k})}{8 \sqrt{3}}\)
= ± \(\frac{1}{\sqrt{3}}(\hat{i}-\hat{j}-\hat{k})\)

Question 15 (old).
(i) If \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k}\), \(\vec{b}=\hat{i}+2 \hat{j}-\hat{k}\) and \(\vec{c}=2 \hat{i}+3 \hat{j}\), find \((\vec{a} \times \vec{b}) \times \vec{c}\) and \(\vec{a} \times(\vec{b} \times \vec{c})\), and verify that these are not the same.
(ii) If \(\vec{a}, \vec{b} \text { and } \vec{c}\) represent the position vectors of the points with coordinates (2, – 10, 2), (3, 1, 2) and (2, 1, 3) respectively, find the value of \(\vec{a} \times(\vec{b} \times \vec{c})\). (ISC 2009)
Solution:
(i) Given \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k}\),
\(\vec{b}=\hat{i}+2 \hat{j}-\hat{k}\)
and \(\vec{c}=2 \hat{i}+3 \hat{j}\)

\(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & -1 & 1 \\
1 & 2 & -1
\end{array}\right|\)
= \(\hat{i}(1-2)-\hat{j}(-2-1)+\hat{k}(4+1)\)
= \(-\hat{i}+3 \hat{j}+5 \hat{k}\)

(ii) Given \(\vec{a}=2 \hat{i}-10 \hat{j}+2 \hat{k}\),
\(\vec{b}=3 \hat{i}+\hat{j}+2 \hat{k}\)
and \(\vec{c}=2 \hat{i}+\hat{j}+3 \hat{k}\)

∴ \(\vec{b} \times \vec{c}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & 1 & 2 \\
2 & 1 & 3
\end{array}\right|\)
= \(\hat{i}(3-2)-\hat{j}(9-4)+\hat{k}(3-2)\)
= \(\hat{i}-5 \hat{j}+\hat{k}\)

∴ \(\vec{a} \times(\vec{b} \times \vec{c})=\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{k} \\
2 & -10 & 2 \\
1 & -5 & 1
\end{array}\right|\)
= \(\hat{i}(-10+10)-\hat{j}(2-2)+\hat{k}(-10+10)\)
= \(0 \hat{i}+0 \hat{j}+0 \hat{k}\)
= \(\overrightarrow{0}\)

Question 16.
(i) Find a vector of magnitude 6 units which is perpendicular to both the vectors \(2 \hat{i}-\hat{j}+2 \hat{k}\) and \(4 \hat{i}-\hat{j}+3 \hat{k}\) (NCERT Exemplar)
(ii) Find a vector of magnitude 19 which is perpendicular to both the vectors \(-\hat{j}+\hat{k}\) and \(4 \hat{i}-\hat{j}+8 \hat{k}\).
Solution:
(i) Let \(\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}\)
and \(\vec{b}=4 \hat{i}-\hat{j}+3 \hat{k}\)

(ii) Let \(\vec{a}=4 \hat{i}-\hat{j}+8 \hat{k}\)
and \(\vec{b}=-\hat{j}+\hat{k}\)

Question 17.
Find a vector of magnitude 6, perpendicular to each of the vectors \((\vec{a}+\vec{b})\) and \((\vec{a}-\vec{b})\) where \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\) and \(\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}\).
Solution:
Given, \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\)
and \(\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}\)

Question 18.
If \(\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}\) and \(\vec{b}=2 \hat{i}+3 \hat{j}-5 \hat{k}\), find \(\vec{a} \times \vec{b}\). Verify that \(\vec{a}\) and \(\vec{a} \times \vec{b}\) are perpendicular to each other.
Solution:
Given \(\vec{a}=\hat{i}-2 \hat{j}+3 \hat{k}\)
and \(\vec{b}=2 \hat{i}+3 \hat{j}-5 \hat{k}\)
\(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -2 & 3 \\
2 & 3 & -5
\end{array}\right|\)
= \(\hat{i}(10-9)-\hat{j}(-5-6)+\hat{k}(3+4)\)
= \(\hat{i}+11 \hat{j}+7 \hat{k}\)

Now \(\vec{a} \cdot(\vec{a} \times \vec{b})\)
= \((\hat{i}-2 \hat{j}+3 \hat{k}) \cdot(\hat{i}+11 \hat{j}+7 \hat{k})\)
= 1 (1) – 2 (11) + 3 (7) = 0
∴ \(\vec{a} \perp \vec{a} \times \vec{b}\)

Question 18 (old).
Find a unit vector perpendicular to the vectors \(4 \hat{i}+3 \hat{j}+\hat{k}\) and \(2 \hat{i}-\hat{j}+2 \hat{k}\). Determine the sine of the angle between these two vectors. (ISC 2007)
Solution:
Given \(\vec{a}=4 \hat{i}+3 \hat{j}+\hat{k}\)
and \(\vec{b}=2 \hat{i}-\hat{j}+2 \hat{k}\)

Question 19.
(i) If \(\vec{a}=\hat{i}-\hat{j}\), \(\vec{b}=3 \hat{j}-\hat{k}\) and \(\vec{c}=7 \hat{i}-\hat{k}\), find a vector \(\vec{d}\) which is perpendicular to both \(\vec{a} \text { and } \vec{b}\) and \(\vec{c} \cdot \vec{d}\) = 1.
(ii) If \(\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}\), \(\vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}\), and \(\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}\), find a vector \(\vec{d}\) which is perpendicular to both the vectors \(\vec{a} \text { and } \vec{b}\) and \(\vec{c} \cdot \vec{d}\) = 27.
(iii) If \(\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}\), \(\vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}\) and \(\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}\). Find a vector \(\vec{d\) which is perpendicular to both \(\vec{a} \text { and } \vec{b}\) and \(\vec{c} \cdot \vec{d}\) = 18.
(iv) If \(\vec{a}=4 \hat{i}+5 \hat{j}-\hat{k}\), \(\vec{b}=\hat{i}-4 \hat{j}+5 \hat{k}\) and \(\vec{c}=3 \hat{i}+\hat{j}-\hat{k}\). Find a vector \(\vec{d\) which is perpendicular to both \(\vec{c} \text { and } \vec{b}\) and \(\vec{d} \cdot \vec{a}\) = 21.
Solution:
(i) Given \(\vec{a}=\hat{i}-\hat{j}\),
\(\vec{b}=3 \hat{j}-\hat{k}\)
and \(\vec{c}=7 \hat{i}-\hat{k}\)

(ii) Given \(\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}\),
\(\vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}\),
and \(\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}\)
Let the required vector be \(\vec{d}=x \hat{i}+y \hat{j}+z \hat{k}\)
such that \(\vec{d}\) is perpendicular to \(\vec{a} \text { and } \vec{b}\)
∴ \(\vec{d} \cdot \vec{b}\) = 0
⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+4 \hat{j}+2 \hat{k})\) = 0
⇒ x + 4y + 7z = 0 …………………(1)
Also \(\vec{d} \cdot \vec{b}\) = 0
⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(3 \hat{i}-2 \hat{j}+7 \hat{k})\) = 0
⇒ 3x – 2y + 7z = 0 ………………..(2)
also given \(\vec{c} \cdot \vec{d}\) = 27
⇒ \((2 \hat{i}-\hat{j}+4 \hat{k}) \cdot(x \hat{i}+y \hat{j}+z \hat{k})\)
⇒ 2x – y + 4z = 0 ………………..(3)
Multiplying eqn. (2) by 2 and adding to eqn. (1) ; we get
7x + 16z = 0 …………………..(4)
Multiplying eqn. (3) by 2 ; we have
4x – 2y + 8z = 54 …………………….(5)
eqn. (5) – eqn. (2) gives ;
x + z = 54 ……………….(6)
Multiplying eqn. (6) by (7) – eqn. (4) ; gives
⇒ 7z – 16z = + 54 × 7
⇒ – 9z = + 54 × 7
⇒ z = – 42
∴ from eqn. (6) ;
x = 96
∴ from eqn. (3) ;
y = 192 – 168 – 27 = – 3
Thus required vector be \(\vec{d}=96 \hat{i}-3 \hat{j}-42 \hat{k}\)
i.e., \(\vec{d}=3(32 \hat{i}-\hat{j}-14 \hat{k})\)

(iii) Given \(\vec{a}=\hat{i}+4 \hat{j}+2 \hat{k}\),
\(\vec{b}=3 \hat{i}-2 \hat{j}+7 \hat{k}\)
and \(\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}\)
Let the required vector be
\(\vec{d}=x \hat{i}+y \hat{j}+z \hat{k}\) …………………..(1)

Since \(\vec{d}\) is perpendicular to \(\vec{a} \text { and } \vec{b}\)
∴ \(\vec{d} \cdot \vec{b}\) = 0
⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+4 \hat{j}+2 \hat{k})\) = 0
⇒ x + 4y + 2z = 0 ………………….(2)

Also \(\vec{d} \cdot \vec{b}\) = 0
⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(3 \hat{i}-2 \hat{j}+7 \hat{k})\) = 0
⇒ 3x – 2y + 7z = 0 ……………………..(3)

given \(\vec{c} \cdot \vec{d}\) = 18
⇒ \((2 \hat{i}-\hat{j}+4 \hat{k}) \cdot(x \hat{i}+y \hat{j}+z \hat{k})\) = 18
⇒ 2x – y + 4z = 18 ……………………..(4)
Multiplying eqn. (3) by (2) and adding to eqn. (2) ; we have
7x + 16z = 0 ……………………(5)
Multiplying eqn. (4) by (2) – eqn. (3) ; we have
x + z = 36 ……………………..(6)
Multiply eqn. (6) by 7 – eqn. (5) ; we have
– 9z = 36 × 7
⇒ z = – 28
∴ from (6) ;
x = 64
∴ from (4) ;
y = 128 – 112 – 18 = – 2
∴ from (1) ;
\(\vec{d}=64 \hat{i}-2 \hat{j}-28 \hat{k}\)
= \(2(32 \hat{i}-\hat{j}-14 \hat{k})\)

(iv) Given \(\vec{a}=4 \hat{i}+5 \hat{j}-\hat{k}\),
\(\vec{b}=\hat{i}-4 \hat{j}+5 \hat{k}\)
and \(\vec{c}=3 \hat{i}+\hat{j}-\hat{k}\)
Let \(\vec{d}=x \hat{i}+y \hat{j}+z \hat{k}\)
Since \(\vec{d}\) is perpendicular to \(\vec{c}\)
⇒ \(\vec{d} \text { and } \vec{c}\) = 0
⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(3 \hat{i}+\hat{j}-\hat{k})\) = 0
⇒ 3x + y – z = 0 ………………..(1)
Also \(\vec{d}\) is perpendicular to \(\vec{b\)
⇒ \(\vec{d} \text { and } \vec{b}\) = 0
i.e. \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}-4 \hat{j}+5 \hat{k})\) = 0
⇒ x – 4y + 5z = 0 …………………..(2)
Also, \(\) = 21
⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(4 \hat{i}+5 \hat{j}-\hat{k})\) = 21
⇒ 4x + 5y – z = 21 ……………………(3)
Multiply eqn. (1) by 5 and adding to eqn. (2) ; we get
16x + y = 0 ………………….(4)
Multiply eqn. (3) by 5 and adding to eqn. (2) ; we get
21x + 21y = 105
⇒ x + y = 5 ……………………(5)
On solving eqn. (4) and (5) ; we have
x = – \(\frac{1}{3}\)
and y = \(\frac{16}{3}\)

Question 20.
Define \(\vec{a} \times \vec{b}\) and prove that \(|\vec{a} \times \vec{b}|=(\vec{a} \cdot \vec{b})\) tan θ, where θ is the angle between \(\vec{a} \text { and } \vec{b}\).
Solution:
The vector product \(\) is a vector quantity whose magnitude is \(|\vec{a}||\vec{b}|\) sin θ where θ is the angle between \(\vec{a} \text { and } \vec{b}\) and whose direction is perpendicular to plane of \(\vec{a} \text { and } \vec{b}\) in such a way that \(\vec{a}, \vec{b}\) and this direction forms a right handed system.
∴ \(\vec{a} \times \vec{b}=|\vec{a}||\vec{b}| \sin \theta \hat{n}\)
where \(\hat{n\) be the unit vector ⊥ to the plane of \(\vec{a} \text { and } \vec{b}\) s.t. \(\vec{a}, \vec{b}, \hat{n}\) forms a right handed system.
Since \(|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}|\) ……………………..(1)
We know that \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta\)
∴ \(|\vec{a}||\vec{b}|=\frac{\vec{a} \cdot \vec{b}}{\cos \theta}\)
∴ from (1) ;
\(|\vec{a} \times \vec{b}|=\frac{\vec{a} \cdot \vec{b}}{\cos \theta}\) sin θ
∴ \(|\vec{a} \times \vec{b}|=(\vec{a} \cdot \vec{b})\) tan θ

Question 21.
In ∆ OAB, \(\overrightarrow{\mathrm{OA}}=3 \hat{i}+2 \hat{j}-\hat{k}\) and \(\overrightarrow{\mathrm{OB}}=\hat{i}+3 \hat{j}+\hat{k}\). Find the area of the triangle OAB.
Solution:
Given \(\overrightarrow{\mathrm{OA}}=3 \hat{i}+2 \hat{j}-\hat{k}\)
and \(\overrightarrow{\mathrm{OB}}=\hat{i}+3 \hat{j}+\hat{k}\)

Question 22.
Using vectors, find the area of the triangle whose vertices are : A (3, – 1, 2), B (1, – 1, – 3), C (4, – 3, 1). (ISC 2020)
Solution:

Question 23.
(i) Find the area of a parallelogram whose diagonals are determined by the vectors \(\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}\) and \(\vec{b}=\hat{i}-3 \hat{j}+4 \hat{k}\). (ISC 2009)
(ii) The vectors \(-2 \hat{i}+4 \hat{j}+4 \hat{k}\) and \(-4 \hat{i}-2 \hat{k}\) represent the diagonals BD and AC of a parallelogram ABCD. Find the area of the parallelogram. (ISC 2006)
Solution:
(i) Given \(\vec{a}=3 \hat{i}+\hat{j}-2 \hat{k}\)
and \(\vec{b}=\hat{i}-3 \hat{j}+4 \hat{k}\)
Now \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & 1 & -2 \\
1 & -3 & 4
\end{array}\right|\)
= \(\hat{i}(4-6)-\hat{j}(12+2)+\hat{k}(-9-1)\)
= \(-2 \hat{i}-14 \hat{j}-10 \hat{k}\)
∴ Required area of ||gm = \(\frac{1}{2}|\vec{a} \times \vec{b}|\)
= \(\frac{1}{2} \sqrt{4+196+100}=5 \sqrt{3}\)

(ii) Given \(\vec{a}=-2 \hat{i}+4 \hat{j}+4 \hat{k}\)
and \(\vec{b}=-4 \hat{i}-2 \hat{k}\)
Now, \(\vec{a} \times \vec{b}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
-2 & 4 & 4 \\
-4 & 0 & -2
\end{array}\right|\)
= \(\hat{i}(-8)-\hat{j}(4+16)+\hat{k}(16)\)
= \(-8 \hat{i}-20 \hat{j}+16 \hat{k}\)
= \(4(-2 \hat{i}-5 \hat{j}+4 \hat{k})\)
∴ Required area of ||gm = \(\frac{1}{2}|\vec{a} \times \vec{b}|\)
= \(\frac{\sqrt{64+400+256}}{2}\)
= \(\frac{\sqrt{720}}{2}\)
= \(\frac{12 \sqrt{5}}{2}\)
= 6√5

Question 24.
The two adjacent sides of a parallelogram are represented by the vectors \(2 \hat{i}-4 \hat{j}-5 \hat{k}\) and \(2 \hat{i}+2 \hat{j}+3 \hat{k}\).
Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram.
Solution:
Given, \(\vec{a}=2 \hat{i}-4 \hat{j}-5 \hat{k}\)
and \(\vec{b}=2 \hat{i}+2 \hat{j}+3 \hat{k}\)

Question 25.
Find the area of the triangle formed by the points A (1, 2, 3), B(2, -1 , 4) and C (4, 5, – 1).
Solution:
Given P.V. of A = \(\hat{i}+2 \hat{j}+3 \hat{k}\) ;
P.V. of B = \(2 \hat{i}-\hat{j}+4 \hat{k}\)
and P.V. of C = \(4 \hat{i}+5 \hat{j}-\hat{k}\)
∴ \(\overrightarrow{\mathrm{AB}}\) = P.V. of B – P.V. of A
= \((2 \hat{i}-\hat{j}+4 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k})\)
= \(\hat{i}-3 \hat{j}+\hat{k}\)

\(\overrightarrow{\mathrm{AB}}\) = P.V. of B – P.V. of A
= \((4 \hat{i}+5 \hat{j}-\hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k})\)
= \(3 \hat{i}+3 \hat{j}-4 \hat{k}\)

∴ \(\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
1 & -3 & 1 \\
3 & 3 & -4
\end{array}\right|\)
= \(\hat{i}(12-3)-\hat{j}(-4-3)+\hat{k}(3+9)\)
= \(9 \hat{i}+7 \hat{j}+12 \hat{k}\)

∴ \(|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|=\sqrt{9^2+7^2+12^2}=\sqrt{274}\)
Thus required area of ∆ABC = \(\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|\)
= \(\frac{\sqrt{274}}{2}\)

Students appreciate clear and concise ML Aggarwal Maths for Class 12 Solutions Chapter 1 Vectors Ex 1.2 that guide them through exercises.

ML Aggarwal Class 12 Maths Solutions Section B Chapter 1 Vectors Ex 1.2

Very short answer type questions (1 to 14) :

Question 1.
(i) If \(\vec{a}\) is a unit vector and \((\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})\) = 15, then find \(|\vec{x}|\).
(ii) If \(\vec{p}\) is a unit vector and \((\vec{x}-\vec{p}) \cdot(\vec{x}+\vec{p})\)= 80, then find \(|\vec{x}|\).
Solution:
(i) Given \((\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})\) = 15

(ii) Given \(\vec{p}\) is a unit vector and \((\vec{x}-\vec{p}) \cdot(\vec{x}+\vec{p})\)= 80

Question 2.
(i) If \((\vec{a}-\vec{b}) \cdot(\vec{a}+\vec{b})\) = 12 and \(|\vec{a}|=2|\vec{b}|\), then find \(|\vec{a}| \text { and }|\vec{b}|\).
(ii) If \(|\vec{a}|\) = 3, \(|\vec{b}|\) = 4 and \(\vec{a} \cdot \vec{b}\) = 1, then find \((\vec{a}-\vec{b})^2\).
(iii) If \(|\vec{a}|\) = 2, \(|\vec{b}|\) = 5 and \(\vec{a} \cdot \vec{b}\) = 8, then find \(|\vec{a}-\vec{b}|\).
(iv) If \(|\vec{a}|\) = 3, \(|\vec{b}|\) = 5 and \(\vec{a} \cdot \vec{b}\) = – 8 then find \(|\vec{a}+\vec{b}|\).
Solution:
(i) Given \((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})\) = 12

(ii) Given \(|\vec{a}|\) = 3, \(|\vec{b}|\) = 4
and \(\vec{a} \cdot \vec{b}\) = 1

(iii) Given \(|\vec{a}|\) = 2, \(|\vec{b}|\) = 5
and \(\vec{a} \cdot \vec{b}\) = 8

(iv) Given \(|\vec{a}|\) = 3, \(|\vec{b}|\) = 5
and \(\vec{a} \cdot \vec{b}\) = – 8

Question 3.
Find the magnitude of each of the two vectors \(\vec{a} \text { and } \vec{b}\), having same magnitude such that the angle between them is 60° and their scalar product is \(\frac{9}{2}\).
Solution:
Given \(|\vec{a}|=|\vec{b}|\) ;
θ = 60° ;
\(\vec{a} \cdot \vec{b}=\frac{9}{2}\)

Question 3 (old).
If the angle between two vectors \(\vec{a} \text { and } \vec{b}\) of equal magnitude is 60° and their scalar product is \(\frac{1}{2}\), then find their magnitudes. (NCERT)
Solution:
Given \(|\vec{a}|=|\vec{b}|\) ……………..(1)
Let θ be the angle between a and b such that θ = 60°
and \(\vec{a} \cdot \vec{b}=\frac{1}{2}\) [using (1)]
We know that
⇒ \(\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta\)
⇒ \(\frac{1}{2}=|\vec{a}| \cdot|\vec{b}| \cdot \cos \frac{\pi}{3}\)
⇒ \(\frac{1}{2}=|\vec{a}|^2 \cdot \frac{1}{2}\)
⇒ \(|\vec{a}|=1=|\vec{b}|\)

Question 4.
(i) If \(|\vec{a}|\) = 2, \(|\vec{b}|\) = √3 and \(\vec{a} \cdot \vec{b}\) = 3, then find the angle between \(\vec{a} \text { and } \vec{b}\).
(ii) Find the angle between the vectors \(\vec{a} \text { and } \vec{b}\) such that \(|\vec{a}|=|\vec{b}|\) = 3 and \(\vec{a} \cdot \vec{b}\) = 1.
(iii) Find the angle between two vectors having the same length √2 and scalar product – 1.
(iv) If \(|\vec{a}|\) = √3, \(|\vec{b}|\) = 2 and the angle between \(\vec{a} \text { and } \vec{b}\) is 60°, find \(\vec{a} \cdot \vec{b}\) = 1.
Solution:
(i) Given \(|\vec{a}|\) = 3, \(|\vec{b}|\) = 2
and \(\vec{a} \cdot \vec{b}\) = 3
Let θ be the angle between \(\vec{a} \text { and } \vec{b}\).
∴ \(\vec{a} \cdot \vec{b}\) = \(|\vec{a}||\vec{b}|\) cos θ
⇒ 3 = 3 × 2 × cos θ
⇒ cos θ = \(\frac{1}{2}\) ; 0 ≤ θ ≤ π
∴ θ = \(\frac{\pi}{3}\)

(ii) Given \(|\vec{a}|\) = 3, \(|\vec{b}|\) = 3 ;
and \(\vec{a} \cdot \vec{b}\) = 1
Let θ be the angle between \(\vec{a} \text { and } \vec{b}\).
∴ cos θ = \(\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}||\vec{b}|}\)
= \(\frac{1}{3 \times 3}=\frac{1}{9}\)
⇒ θ = cos^-1 (\(\frac{1}{9}\))

(iii) Let θ be the angle between two vectors \(\vec{a} \text { and } \vec{b}\) such that
\(|\vec{a}|=|\vec{b}|\) = √2
and \(\vec{a} \cdot \vec{b}\) = – 1
Since cos θ = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\)
= \(\frac{-1}{\sqrt{2} \times \sqrt{2}}=-\frac{1}{2}\)
⇒ cos θ = – cos \(\frac{\pi}{3}\)
= cos (180° – 60°)
= cos 120°
∴ θ = 120°

(iv) Given \(|\vec{a}|\) = √3 , \(|\vec{b}|\) = 2 ;
and θ = 60°
where θ be the angle between \(\vec{a} \text { and } \vec{b}\)
∴ \(\vec{a} \cdot \vec{b}\) = \(|\vec{a}||\vec{b}|\) cos θ
= √3 × 2 × cos 60°
= 2√3 × \(\frac{1}{2}\)
= √3

Question 4 (old).
(i) If \(|\vec{a}|\) = 3, \(|\vec{b}|\) = 2 and \(\vec{a} \cdot \vec{b}\) = 3, then find the angle between \(\vec{a} \text { and } \vec{b}\).
(iv) Find the angle between the vectors \(\vec{a} \text { and } \vec{b}\) with magnitudes 1 and 2 respectively, and \(\vec{a} \cdot \vec{b}\) = 1
Solution:
(ii) Given \(|\vec{a}|\) = 2, \(|\vec{b}|\) = √3
and \(\vec{a} \cdot \vec{b}\) = 3
Let θ be the angle between \(\vec{a} \text { and } \vec{b}\).
Then \(\vec{a} \cdot \vec{b}\) = \(|\vec{a}||\vec{b}|\) cos θ
⇒ 3 = 2 × √3 × cos θ
⇒ cos θ = \(\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2}\) ; 0 ≤ θ ≤ π
∴ θ = 30°.

(iv) Given \(|\vec{a}|\) = 1, \(|\vec{b}|\) = 2 ;
and \(\vec{a} \cdot \vec{b}\) = 1
Let θ be the angle between \(\vec{a} \text { and } \vec{b}\).
∴ cos θ = \(\frac{(\vec{a} \cdot \vec{b})}{|\vec{a}||\vec{b}|}\)
= \(\frac{1}{1 \cdot 2}=\frac{1}{2}\)
⇒ θ = \(\frac{\pi}{3}\)

Question 5.
Find the angle between the vectors \(2 \hat{i}-\hat{j}+\hat{k}\) and \(3 \hat{i}+4 \hat{j}-\hat{k}\). (NCERT Exemplar)
Solution:
Let \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k}\)
and \(\vec{b}=3 \hat{i}+4 \hat{j}-\hat{k}\)

Question 5 (old).
(i) Find the angle between the vectors \(\vec{a}=\hat{i}+\hat{j}-\hat{k}\) and \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\).
Solution:
Given \(\vec{a}=\hat{i}+\hat{j}-\hat{k}\)
and \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\)

Question 6.
Find
(i) \((\vec{b}-\vec{a}) \cdot(3 \vec{a}+\vec{b})\) where \(\vec{a}=\hat{i}-2 \hat{j}+5 \hat{k}\), \(\overrightarrow{\vec{b}}=2 \hat{i}+\hat{j}-3 \hat{k}\). \((\vec{a}+3 \vec{b}) \cdot(2 \vec{a}-\vec{b})\)
(ii) where \(\vec{a}=\hat{i}+\hat{j}+2 \hat{k}\), \(\vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}\)
Solution:
(i) Given \(\vec{a}=\hat{i}-2 \hat{j}+5 \hat{k}\),
and \(\overrightarrow{\vec{b}}=2 \hat{i}+\hat{j}-3 \hat{k}\)

(ii) Given \(\vec{a}=\hat{i}+\hat{j}+2 \hat{k}\)
and \(\vec{b}=3 \hat{i}+2 \hat{j}-\hat{k}\)

Question 7.
(i) Find the projection of \(\vec{a} \text { on } \vec{b}\) if \(\vec{a} \cdot \vec{b}\) = 8 and \(\vec{b}=2 \hat{i}+6 \hat{j}+3 \hat{k}\).
(ii) Find the projection of \(\hat{i}+3 \hat{j}+7 \hat{k}\) on the vector \(2 \hat{i}-3 \hat{j}+6 \hat{k}\).
(iii) Find the projection of \(\hat{i}-\hat{j}\) on the vector \(\hat{i}+\hat{j}\).
(iv) Find λ when the projection of \(\hat{i}+\lambda \hat{j}+\hat{k}\) on \(\hat{i}+\hat{j}\) is √2 units.
Solution:
(i) Given \(\vec{a} \cdot \vec{b}\) = 8
and \(\vec{b}=2 \hat{i}+6 \hat{j}+3 \hat{k}\)
∴ \(|\vec{b}|=\sqrt{2^2+6^2+3^2}\)
= \(\sqrt{49}\) = 7
Thus required projection of \(\vec{a} \text { on } \vec{b}\)
= \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=\frac{8}{7}\)

(ii) Let \(\vec{a}=\hat{i}+3 \hat{j}+7 \hat{k}\)
and \(\vec{b}=2 \hat{i}-3 \hat{j}+6 \hat{k}\)
∴ \(\vec{a} \cdot \vec{b}=(\hat{i}+3 \hat{j}+7 \hat{k}) \cdot(2 \hat{i}-3 \hat{j}+6 \hat{k})\)
= 1 (2) + 3 (- 3) + 7 (6)
= 2 – 9 + 42 = 35
and \(|\vec{b}|=\sqrt{2^2+(-3)^2+6^2}\)
= \(\sqrt{49}\) = 7
∴ required projection of \(\vec{a} \text { on } \vec{b}\) = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\)
= \(\frac{35}{7}\) = 5

(iii) Given \(\vec{a}=\hat{i}-\hat{j}\)
and \(\vec{b}=\hat{i}+\hat{j}\)
∴ \(\vec{a} \cdot \vec{b}\) = 1 (1) – (1) (1) = 0
∴ scalar projection of \(\vec{a} \text { on } \vec{b}\) = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\) = 0

(iv) Let \(\vec{a}=\hat{i}+\lambda \hat{j}+\hat{k}\)
and \(\vec{b}=\hat{i}+\hat{j}\)
∴ \(\vec{a} \cdot \vec{b}=(\hat{i}+\lambda \hat{j}+\hat{k}) \cdot(\hat{i}+\hat{j}+0 \hat{k})\)
= 1 (1) + λ (1) + 1 (0)
= 1 + λ
\(|\vec{b}|=\sqrt{1^2+1^2}=\sqrt{2}\)
We know that projection \(\vec{a} \text { on } \vec{b}\) = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\)
∴ √2 = \(\frac{1+\lambda}{\sqrt{2}}\)
⇒ 1 + λ = 2
⇒ λ = 1.

Question 8.
(i) If \(\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}\) and \(\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k}\), then show that the vectors \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\) are perpendicular to each other. (NCERT)
(ii) Find λ if \(\vec{a}=3 \hat{i}-\hat{j}+4 \hat{k}\) and \(\vec{b}=-\lambda \hat{i}+3 \hat{j}+3 \hat{k}\) are perpendicular to each other.
(iii) For what value of λ are the vectors \(\vec{a}=2 \hat{i}+\lambda \hat{j}+\hat{k}\) and \(\vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}\) perpendicular to each other ?
Solution:
(i) Given \(\vec{a}=5 \hat{i}-\hat{j}-3 \hat{k}\)
and \(\vec{b}=\hat{i}+3 \hat{j}-5 \hat{k}\)

(ii) Given \(\vec{a}=3 \hat{i}-\hat{j}+4 \hat{k}\)
and \(\vec{b}=-\lambda \hat{i}+3 \hat{j}+3 \hat{k}\)
Since \(\vec{a}\) is ⊥ to \(\vec{b}\)
∴ \(\vec{a} \cdot \vec{b}\) = 0
⇒ \((3 \hat{i}-\hat{j}+4 \hat{k}) \cdot(-\lambda \hat{i}+3 \hat{j}+3 \hat{k})\) = 0
⇒ 3 (- λ) – 1 (3) + 4 (3) = 0
⇒ – 3λ + 9 = 0
⇒ λ = 3.

(iii) Given \(\vec{a}=2 \hat{i}+\lambda \hat{j}+\hat{k}\)
and \(\vec{b}=\hat{i}-2 \hat{j}+3 \hat{k}\)
Now \(\vec{a} \text { and } \vec{b}\) are orthogonal or perpendicualr.
∴ \(\vec{a} \cdot \vec{b}\) = 0
⇒ 2 (1) + λ (- 2) + 1 (3) = 0
⇒ 2λ = 5
⇒ λ = \(\frac{5}{2}\)

Question 9.
(i) If \(\vec{a} \text { and } \vec{b}\) are unit vectors, then what is the angle between \(\vec{a} \text { and } \vec{b}\) so that \(\sqrt{3} \vec{a}-\vec{b}\) may be a unit vector. (NCERT Exemplar)
(ii) If \(|\vec{a}|\) = 2, \(|\vec{b}|\) = 3 and \(\vec{a} \cdot \vec{b}\) = 4, then what is the value of \(|\vec{a}+2 \vec{b}|\) ?
(iii) If \(|\vec{a}|\) = 2, \(|\vec{b}|\) = 3 and \(\vec{a} \cdot \vec{b}\) = 4, find \(|2 \vec{a}-3 \vec{b}|\).
(iv) If \(|\vec{a}|\) = 2, \(|\vec{b}|\) = 3 and \(|2 \vec{a}-\vec{b}|\) = 4, then find \(|2 \vec{a}+\vec{b}|\).
Solution:
(i) Given \(\vec{a} \text { and } \vec{b}\) are unit vectors
∴ \(|\vec{a}|\) = \(|\vec{b}|\) = 1
since \(\sqrt{3} \vec{a}-\vec{b}\) is a unit vector

(ii) Given \(|\vec{a}|\) = 2, \(|\vec{b}|\) = 3
and \(\vec{a} \cdot \vec{b}\) = 4

= 2² + 4 (4) + 4 (9)
= 4 + 16 + 36 = 56
∴ \(|\vec{a}+2 \vec{b}|=\sqrt{56}=2 \sqrt{14}\)

(iii) Given \(|\vec{a}|\) = 2, \(|\vec{b}|\) = 3
and \(\vec{a} \cdot \vec{b}\) = 4

(iv) Given \(|\vec{a}|\) = 2,
\(|\vec{b}|\) = 3
and \(|2 \vec{a}-\vec{b}|\) = 4

Question 10.
If \(\vec{a}, \vec{b}\) are unit vectors and c = \(|\vec{a}+\vec{b}|\), d = \(|\vec{a}-\vec{b}|\), then what is the value of c² + d² ?
Solution:
Given c = \(|\vec{a}+\vec{b}|\)

Question 11.
If \(\vec{a}, \vec{b}, \vec{c}\) are three vectors such that \(\vec{c}=\vec{a}+\vec{b}\) and \(\vec{a} \cdot \vec{b}\) = 0, then show that c² = a² + b².
Solution:
Given \(\vec{c}=\vec{a}+\vec{b}\)

Question 12.
(i) If \(\vec{a}, \vec{b} \text { and } \vec{c}\) are unit vectors such that \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\), then find the value of \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}\). (NCERT)
(ii) If \(\vec{a}, \vec{b}, \vec{c}\) are three vectors \(|\vec{a}|\) = 5, \(|\vec{b}|\) = 12, \(|\vec{c}|\) = 13 and \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}\) then find the value of \(\).
Solution:
(i) Given \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\) ;
on squaring, we ahve

(ii) Given \(|\vec{a}|\) = 5 ;
\(|\vec{b}|\) = 12 ;
\(|\vec{c}|\) = 13
and \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\) ;
on squaring we have
∴

Question 13.
(i) We are given two vectors \(\vec{a} \text { and } \vec{b}\) such that \((\vec{a})^2=(\vec{b})^2\). Is it necessary that \(\vec{a}=\vec{b}\) ? Justify your answer.
(ii) For two non-zero vectors \(\vec{a} \text { and } \vec{b}\), state when \(|\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2\) holds.
(iii) If \(\vec{a}=\overrightarrow{0}\) or \(\vec{b}=\overrightarrow{0}\), then \(\vec{a} \cdot \vec{b}\) = 0. Is the converse true ? Justify your answer by an example.
(iv) If \(\vec{a}, \vec{b}\) are non – zero vectors and \(\vec{a} \cdot \vec{b}\) ≥ 0, then what can you say about the angle θ between the vectors \(\vec{a} \text { and } \vec{b}\).
Solution:
(i) Let \(\vec{a}=\hat{i}+2 \hat{j}\)
and \(\vec{b}=2 \hat{i}-\hat{j}\)

Here \(\vec{a}^2=|\vec{a}|^2\)
= \(\left(\sqrt{1^2+2^2}\right)^2\)
= \((\sqrt{5})^2\) = 5

and \(\vec{b}^2=|\vec{b}|^2\)
= \(\left(\sqrt{2^2+(-1)^2}\right)^2\)
= \((\sqrt{5})^2\) = 5
i.e. \(\vec{a}^2=\vec{b}^2\) but \(\vec{a} \neq \vec{b}\).

(ii) \(|\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2\) holds

(iii) \(\vec{a} \cdot \vec{b}=\overrightarrow{0} \cdot \vec{b}\) = 0
if \(\vec{a}\) = 0
When \(\vec{b}\) = 0 ;
\(\vec{a} \cdot \vec{b}=\vec{a} \cdot \overrightarrow{0}\) = 0
but converse need not be true.
e.g. \(\vec{a}=2 \hat{i}-\hat{j}+\hat{k}\)
and \(\vec{b}=\hat{i}+\hat{j}-\hat{k}\)
but \(\vec{a} \cdot \vec{b}=(2 \hat{i}-\hat{j}+\hat{k}) \cdot(\hat{i}+\hat{j}-\hat{k})\)
= 2 (1) – 1 (1) + 1 (- 1)
= 2 – 2 = 0
Here \(\vec{a} \cdot \vec{b}\) = 0 but neither \(\vec{a}=\overrightarrow{0}\) nor \(\vec{b}=\overrightarrow{0}\).

(iv) Since \(\vec{a} \text { and } \vec{b}\) are non-zero vectors.
∴ \(|\vec{a}| \cdot \mid \vec{b}\) > 0
also \(\vec{a} \cdot \vec{b}\) ≥ 0
Thus cos θ = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}| \cdot|\vec{b}|}\) ≥ 0
∴ 0 ≤ θ ≤ \(\frac{\pi}{2}\)

Question 14.
If \(\vec{a} \cdot \vec{a}\) = 0 and \(\vec{a} \cdot \vec{b}\) = 0, then what can you say about the vector \(\vec{b}\) ?
Solution:
Given \(\vec{a} \cdot \vec{a}\) = 0
and \(\vec{a} \cdot \vec{b}\) = 0
⇒ \(|\vec{a}|^2\) = 0 i.e. \(\vec{a}=\overrightarrow{0}\)
and (\(\vec{a} \perp \vec{b}\) or \(\vec{a}=\overrightarrow{0}\) or \(\vec{b}=\overrightarrow{0}\))
i.e. \(\vec{a}=\overrightarrow{0}\) and \(\vec{b}\) can be any vector.

Question 15.
Find the angle between the vectors \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\) if \(\vec{a}=2 \hat{i}-\hat{j}+3 \hat{k}\) and \(\vec{b}=3 \hat{i}+\hat{j}-2 \hat{k}\).
Solution:
Given \(\vec{a}=2 \hat{i}-\hat{j}+3 \hat{k}\)
and \(\vec{b}=3 \hat{i}+\hat{j}-2 \hat{k}\)

Question 16.
Find the projection of \(\vec{b}+\vec{c}\) on \(\vec{a}\) where \(\vec{a}=2 \hat{i}-2 \hat{j}+\hat{k}\), \(\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}\) and \(\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}\).
Solution:
Given \(\vec{a}=2 \hat{i}-2 \hat{j}+\hat{k}\),
\(\vec{b}=\hat{i}+2 \hat{j}-2 \hat{k}\)
and \(\vec{c}=2 \hat{i}-\hat{j}+4 \hat{k}\)

Here \(\vec{b} \cdot \vec{a}=(2 \hat{i}-2 \hat{j}+\hat{k}) \cdot(\hat{i}+2 \hat{j}-2 \hat{k})\)
= 2 (1) – 2 (2) + 1 (- 2)
= 2 – 4 – 2 = – 4

\(\vec{c} \cdot \vec{a}=(2 \hat{i}-\hat{j}+4 \hat{k}) \cdot(2 \hat{i}-2 \hat{j}+\hat{k})\)
= 2 (2) – 1 (- 2) + 4 (1)
= 4 + 2+ 4 = 10

∴ required projection of \(\vec{b}+\vec{c} \text { on } \vec{a}\)
= \(\frac{(\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}|}\)
= = \(\frac{(\vec{b} \cdot \vec{a})+(\vec{c} \cdot \vec{a})}{|\vec{a}|}\)
= \(\frac{-4+10}{\sqrt{2^2+(-2)^2+1^2}}\)
= \(\frac{6}{3}\)
= 2

Question 17.
(i) If \(\vec{a}=\hat{i}-\hat{j}+7 \hat{k}\) and \(\vec{b}=5 \hat{i}-\hat{j}+\lambda \hat{k}\), then find λ such that \(\vec{a}-\vec{b}\) and \(\vec{a}+\vec{b}\) are perpendicular to each other.
(ii) For any (non-zero) vectors \(\vec{a}\), and \(\vec{b}\) prove that the vectors \(|\vec{a}| \vec{b}+|\vec{b}| \vec{a}\) and \(|\vec{a}| \vec{b}-|\vec{b}| \vec{a}\) are perpendicular to each other. (NCERT)
Solution:
(i) Given \(\vec{a}=\hat{i}-\hat{j}+7 \hat{k}\)
and \(\vec{b}=5 \hat{i}-\hat{j}+\lambda \hat{k}\)
Since \(\vec{a}-\vec{b}\) and \(\vec{a}-\vec{b}\) are ⊥ to each other.
∴ \((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})\) = 0
⇒ \(\vec{a} \cdot \vec{a}-\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{a}-\vec{b} \cdot \vec{b}\) = 0
⇒ \(|\vec{a}|^2-|\vec{b}|^2\) = 0
[∵\(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}\)]
⇒ {1² + (- 1)² + 7²} – {5² + (- 1)² + λ²} = 0
⇒ 51 – (26 + λ²) = 0
⇒ 25 – λ² = 0
⇒ λ = ± 5

(ii) Now,

Question 18.
If \(\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}\), \(\vec{b}=-\hat{i}+2 \hat{j}+\hat{k}\) and \(\vec{c}=3 \hat{i}+\hat{j}\) are such that \(\vec{a}+\lambda \vec{b}\) is perpendiculai to \(\vec{c}\), then find the value of λ ?
Solution:
Given \(\vec{a}=2 \hat{i}+2 \hat{j}+3 \hat{k}\),
\(\vec{b}=-\hat{i}+2 \hat{j}+\hat{k}\)
and \(\vec{c}=3 \hat{i}+\hat{j}\)

Now \(\vec{a}+\lambda \vec{b}\)
= \((2 \hat{i}+2 \hat{j}+3 \hat{k})+\lambda(-\hat{i}+2 \hat{j}+\hat{k})\)
= \((2-\lambda) \hat{i}+(2+2 \lambda) \hat{j}+(3+\lambda) \hat{k}\)

Now \(\vec{a}+\lambda \vec{b}\) is ⊥ to \(\vec{c}\)
∴ \(\vec{a}+\lambda \vec{b}\) . \(\vec{c}\) = 0
i.e. \(\) = 0
⇒ 3 (2 – λ) + (2 + 2λ) 1 = 0
⇒ 8 – λ = 0
⇒ λ = 8

Question 19.
(i) If \(\vec{c}\) is perpendicular to \(\vec{a} \text { and } \vec{b}\) both, then show that it is perpendicular to \(\vec{a}+\vec{b}\) as well as \(\vec{a}-\vec{b}\).
(ii) Show that the vectors \(\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k})\), \(\frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k})\) and \(\frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})\) are mutually perpendicular unit vccotrs. (NCERT)
(iii) If \(\vec{a} \text { and } \vec{b}\) are unit vectors and θ is the angle between them, then prove that cos \(\frac{\theta}{2}=\frac{1}{2}|\vec{a}+\vec{b}|\).
Solution:
(i) Since \(\vec{c}\) is ⊥ to \(\vec{a} \text { and } \vec{b}\)
∴ \(\vec{c} \cdot \vec{a}\) = \(\vec{c} \cdot \vec{b}\)
Now \(\vec{c} \cdot(\vec{a}+\vec{b})=\vec{c} \cdot \vec{a}+\vec{c} \cdot \vec{b}\)
= 0 + 0 = 0
and \(\vec{c} \cdot(\vec{a}-\vec{b})=\vec{c} \cdot \vec{a}-(\vec{c} \cdot \vec{b})\)
= 0 – 0 = 0
Thus \(\vec{c}\) is ⊥ to both \(\vec{a}+\vec{b} \text { and } \vec{a}-\vec{b}\).

(ii) Given \(\vec{a}=\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k})\),
\(\vec{b}=\frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k})\)
and \(\vec{c}=\frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})\)

(iii) Given \(\vec{a} \text { and } \vec{b}\) s.t. \(|\hat{a}|=|\hat{b}|\) = 1
and given θ be the angle between them
Now, \(|\hat{a}+\hat{b}|^2=(\hat{a}+\hat{b}) \cdot(\hat{a}+\hat{b})\)
= \(|\hat{a}|^2+|\hat{b}|^2+2 \hat{a} \cdot \hat{b}\)
[∵ \(\hat{a}^2=|\vec{a}|^2, \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{a}\)]
= 1 + 1 + 2 \(|\hat{a}|=|\hat{b}|\) cos θ
= 2 + 2 cos θ
= 2 (1 + cos θ)
= 2 × 2 cos² \(\frac{\theta}{2}\)
= 4 cos² \(\frac{\theta}{2}\)
⇒ \(|\hat{a}+\hat{b}|=2 \cos \frac{\theta}{2}\)
\(\cos \frac{\theta}{2}=\frac{1}{2}|\hat{a}+\hat{b}|\)

Question 20.
Find the angles which the vector \(\vec{a}=3 \hat{i}-6 \hat{j}+2 \hat{k}\) makes with the co-ordinate axes.
Solution:
Given \(\vec{a}=3 \hat{i}-6 \hat{j}+2 \hat{k}\)
Let α, β and γ are the angle made by the vector with the coordinate axes.

Question 21.
(i) Dot products of a vector with vectors \(\), \(\) and \(\) are respectively – 1, 6 and 5. Find the vector.
(ii) The dot product of a vector with the vectors \(\), \(\)and \(\) are 0, 5 and 8 respectively. Find the vector.
Solution:
(i) Let the required vector be
\(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\) …………………..(1)
Now \(\vec{a} \cdot(3 \hat{i}-5 \hat{k})\) = – 1
⇒ 3a₁ – 5a₃ = – 1 ………………………..(2)
also \(\vec{a} \cdot(2 \hat{i}+7 \hat{j})\) = 6
⇒ 2a₁ + 7a₂ = 6 ……………………….(3)
and \(\vec{a} \cdot(\hat{i}+\hat{j}+\hat{k})\) = 5
⇒ a₁ + a₂ + a₃ = 5 ………………(4)
eqn. (3) – 7 × eqn. (4) ; we have
– 5a₁ – 7a₃ = – 29
⇒ 5a₁ + 7a₃ = 29 ………………………(5)
On solving (2) and (5) we get
46a₁ = 145 – 7 = 138
⇒ a₁ = 3 ;
a₃ = 2
∴ from (4) ;
3 + 2 + a₂ = 5
⇒ a₂ = 0

(ii) Let the required vector be
\(\vec{a}=x \hat{i}+y \hat{j}+z \hat{k}\) ……………..(1)
given, \(\vec{a} \cdot(\hat{i}+\hat{j}-3 \hat{k})\) = 0

⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+\hat{j}-3 \hat{k})\) = 0
⇒ x + y – 3z = 0 ………………..(2)
also, \(\vec{a} \cdot(\hat{i}+3 \hat{j}-2 \hat{k})\) = 5

⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(\hat{i}+3 \hat{j}-2 \hat{k})\) = 5
⇒ x + 3y – 2z = 5 ……………….(3)
also, \(\vec{a} \cdot(2 \hat{i}+\hat{j}+4 \hat{k})\) = 8

⇒ \((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(2 \hat{i}+\hat{j}+4 \hat{k})\) = 8
⇒ 2x + y + 4z = 8 ………………(4)

eqn. (3) – eqn. (2) gives ;
2y + z = 5 ………………….(5)
eqn. (4) – 2 × eqn. (3) gives ;
– 5y + 8z = – 2 ……………………(6)
eqn. (6) – 8 × eqn. (5) gives ;
– 5y – 16y = – 2 – 40
⇒ – 21 y = – 42
⇒ y = 2
∴ from(5) ;
4 + z = 5
⇒ z = 1
∴ from (2) ;
x + 2 – 3 = 0
⇒ x = 1
∴ required vector be \(\vec{a}=\hat{i}+2 \hat{j}+\hat{k}\).

Question 22.
Show that the vectors \(\vec{a}=\hat{i}-3 \hat{j}-5 \hat{k}\), \(\vec{b}=2 \hat{i}-\hat{j}+\hat{k}\) and \(\vec{c}=\hat{i}+2 \hat{j}+6 \hat{k}\) form a right-angled triangle.
Solution:
Given \(\vec{a}=\hat{i}-3 \hat{j}-5 \hat{k}\),
\(\vec{b}=2 \hat{i}-\hat{j}+\hat{k}\)
and \(\vec{c}=\hat{i}+2 \hat{j}+6 \hat{k}\)

Now \(\vec{a}+\vec{c}=(\hat{i}-3 \hat{j}-5 \hat{k})+(\hat{i}+2 \hat{j}+6 \hat{k})\)
= \(2 \hat{i}-\hat{j}+\hat{k}=\vec{b}\)

∴ \(\vec{a}, \vec{b} \text { and } \vec{c}\) are coplanar.
also no two of these vectors are parallel.
∴ gives vectors forms a triangle.

Also \(\vec{a} \cdot \vec{b}=(\hat{i}-3 \hat{j}-5 \hat{k}) \cdot(2 \hat{i}-\hat{j}+\hat{k})\)
= 1 (2) – 3 (- 1) – 5 (1) = 0
These dot product of two non-zero vectors is zero.
Therefore these vectors \(\vec{a} \text { and } \vec{b}\) are ⊥ to each other.
Hence the sides represented by these vectors are perpendicular.
Thus, the given vectors forms a right angled triangle.

Question 23.
If A, B, C have position vectors \(\hat{j}+\hat{k}\), \(3 \hat{i}+\hat{j}+5 \hat{k}\) and \(3 \hat{j}+3 \hat{k}\) respectively, then prove that ∆ABC is right angled at C.
Solution:
Let \(\vec{a}=\hat{j}+\hat{k}\),
\(\vec{b}=3 \hat{i}+\hat{j}+5 \hat{k}\)
and \(\vec{c}=3 \hat{j}+3 \hat{k}\)
are the position vectors of points A, B, C respectively.

Question 24.
If the vertices of a AABC are A(- 1, 3, 2), B (2, 3, 5) and C (3, 5, – 2), then show that it is right angled at A. Also find the other two angles.
Solution:
The position vectors of the vertices A, B and C of a triangle are \(-\hat{i}+3 \hat{j}+2 \hat{k}\), \(2 \hat{i}+3 \hat{j}+5 \hat{k}\) and \(3 \hat{i}+5 \hat{j}-2 \hat{k}\)
‘
Hence, the three sides of the triangle are represented by \(\overrightarrow{\mathrm{AB}}, \overrightarrow{\mathrm{BC}} \text { and } \overrightarrow{\mathrm{CA}}\).

∴ \(\overrightarrow{\mathrm{AB}}\) = P.V 0f B – P.V 0f A
= \((2 \hat{i}+3 \hat{j}+5 \hat{k})-(-\hat{i}+3 \hat{j}+2 \hat{k})\)
= \(3 \hat{i}+3 \hat{k}\)

\(\overrightarrow{\mathrm{BC}}\) = P.V 0f C – P.V 0f B
= \((3 \hat{i}+5 \hat{j}-2 \hat{k})-(2 \hat{i}+3 \hat{j}+5 \hat{k})\)
= \(\hat{i}+2 \hat{j}-7 \hat{k}\)

\(\overrightarrow{\mathrm{CA}}\) = P.V 0f A – P.V 0f C
= \((-\hat{i}+3 \hat{j}+2 \hat{k})-(3 \hat{i}+5 \hat{j}-2 \hat{k})\)
= \(-4 \hat{i}-2 \hat{j}+4 \hat{k}\)

∴ \(\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{CA}}\) = \((3 \hat{i}+3 \hat{k}) \cdot(-4 \hat{i}-2 \hat{j}+4 \hat{k})\)
= 3 (- 4) + 0 (- 2) + 3 (4) = 0

Thus \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{CA}}\) are ⊥to each other.
∴ ∠A = 90
Therefore, ∆ ABC is right-angled at A.

Question 25.
Prove that two proper vectors \(\vec{a} \text { and } \vec{b}\) are at right angles iff \((\vec{a}+\vec{b}) \cdot(\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2\). (NCERT)
Solution:

Question 26.
If the coordinates of four points are A (2, 3, 4) , B (5, 4, – 1), C (3, 6, 2) and D (1, 2, 0), then show that \(\overrightarrow{\mathrm{AB}}\) is perpendiculat to \(\overrightarrow{\mathrm{CD}}\). (NCERT Exemplar)
Solution:
Given P.V. of A = \(2 \hat{i}+3 \hat{j}+4 \hat{k}\) ;
P.V. of B = \(5 \hat{i}+4 \hat{j}-\hat{k}\) ;
P.V. of C = \(3 \hat{i}+6 \hat{j}+2 \hat{k}\)
and P.V. of D = \(\hat{i}+2 \hat{j}+0 \hat{k}\)

∴ \(\overrightarrow{\mathrm{AB}}\) = P.V. of B – P.V. of A
= \((5 \hat{i}+4 \hat{j}-\hat{k})-(2 \hat{i}+3 \hat{j}+4 \hat{k})\)
= \(3 \hat{i}+\hat{j}-5 \hat{k}\)

∴ \(\overrightarrow{\mathrm{CD}}\) = P.V. of D – P.V. of C
= \((\hat{i}+2 \hat{j}+0 \hat{k})-(3 \hat{i}+6 \hat{j}+2 \hat{k})\)
= \(-2 \hat{i}-4 \hat{j}-2 \hat{k}\)

Here \(\overrightarrow{\mathrm{AB}} \cdot \overrightarrow{\mathrm{CD}}\)
= \((3 \hat{i}+\hat{j}-5 \hat{k}) \cdot(-2 \hat{i}-4 \hat{j}-2 \hat{k})\)
= 3 (- 2) + 1 (- 4) – 5 (- 2)
= – 6 – 4 + 10 = 0

Question 27.
If \(\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}\) and \(\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}\), then find the angle between \(2 \vec{a}+\vec{b}\) and \(\vec{a}+2 \vec{b}\).
Solution:
Given \(\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}\)
and \(\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}\)

Question 28.
If the points A, B and C with position vectors \(2 \hat{i}+\hat{j}+\hat{k}\), \(\hat{i}-3 \hat{j}-5 \hat{k}\) and \(\alpha \hat{i}-3 \hat{j}+\hat{k}\) respectively are the vertices of a right-angled triangle at C, then find the value (s) of a.
Solution:
Since the points A, B and C with position vectors \(2 \hat{i}+\hat{j}+\hat{k}\), \(\hat{i}-3 \hat{j}-5 \hat{k}\) and \(\alpha \hat{i}-3 \hat{j}+\hat{k}\) are the vertices of right angled triangle at C.

Question 29.
Let \(\vec{a} \text { and } \vec{b}\) be unit vectors. If the vectors \(\vec{c}=\vec{a}+2 \vec{b}\) and \(\vec{d}=5 \vec{a}-4 \vec{b}\) are perpendicualr to each other, then find the angle between \(\vec{a} \text { and } \vec{b}\).
Solution:
Since \(\vec{a} \text { and } \vec{b}\) are unit vectors

Question 30.
Express the vector \(\vec{a}=5 \hat{i}-2 \hat{j}+5 \hat{k}\) as the sum of two vectors such that one is parallel to the vector \(\vec{b}=3 \hat{i}+\hat{k}\) and the other is perpendicular to \(\).
Solution:
We want to find the vectors \(\vec{c} \text { and } \vec{d}\) such that
\(\vec{a}=\vec{c}+\vec{d}\) …………………………..(1)
given \(\vec{a}=5 \hat{i}-2 \hat{j}+5 \hat{k}\)
Since \(\vec{c}\) is parallel to \(\vec{a}\)
∴ \(\vec{c}=\lambda \vec{b}\) for some scalar λ
⇒ \(\vec{c}=\lambda(3 \hat{i}+\hat{k})\)
∴ From (1) ; we have
\(\vec{d}=\vec{a}-\vec{c}\)
= \((5 \hat{i}-2 \hat{j}+5 \hat{k})-(3 \lambda \hat{i}+\lambda \hat{k})\)
⇒ \(\vec{d}=(5-3 \lambda) \hat{i}-2 \hat{j}+(5-\lambda) \hat{k}\)
also it is given that \(\vec{d} \perp \vec{b}\)
⇒ \(\vec{d} \cdot \vec{b}\) = 0
\([(5-3 \lambda) \hat{i}-2 \hat{j}+(5-\lambda) k] \cdot(3 \hat{i}+\hat{k})\) = 0
3 (5 – 3λ) – 2 (0) + (5 – λ) . 1 = 0
15 – 9λ + 5 – λ = 0
⇒ 10λ = 20
⇒ λ = 2.
Thus \(\vec{c}=2(3 \hat{i}+\hat{k})=5 \hat{i}+2 \hat{k}\)
and \(\vec{d}=[(5-6) \hat{i}-2 \hat{j}+(5-2) \hat{k}]\)
= \(-\hat{i}-2 \hat{j}+3 \hat{k}\)

Question 31.
If \(\vec{\alpha}=3 \hat{i}+4 \hat{j}+5 \hat{k}\) and \(\vec{\beta}=2 \hat{i}+\hat{j}-4 \hat{k}\) then express \(\vec{\beta}\) in the form \(\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2\). where \(\overrightarrow{\beta_1}\) is parallel to \(\vec{\alpha} \text { and } \overrightarrow{\beta_2}\) is perpendicualr to \(\vec{\alpha}\).
Solution:
Given \(\vec{\alpha}=3 \hat{i}+4 \hat{j}+5 \hat{k}\)
and \(\vec{\beta}=2 \hat{i}+\hat{j}-4 \hat{k}\)
Given \(\vec{\beta}=\vec{\beta}_1+\vec{\beta}_2\) ……………….(1)
given \(\overrightarrow{\beta_1}\) is parallel to \(\vec{\alpha}\)
∴ \(\overrightarrow{\beta_1}=\lambda \vec{\alpha}\) for some scalar λ.

Question 32.
If the vectors \(\vec{a}=\hat{i}-\hat{j}+2 \hat{k}\), \(\vec{b}=2 \hat{i}+4 \hat{j}+\hat{k}\) and \(\vec{c}=\lambda \hat{i}+\hat{j}+\mu \hat{k}\) are mutually orthogonal, then find the values of λ and μ.
Solution:
Given \(\vec{a}=\hat{i}-\hat{j}+2 \hat{k}\)
\(\vec{b}=2 \hat{i}+4 \hat{j}+\hat{k}\)
and \(\vec{c}=\lambda \hat{i}+\hat{j}+\mu \hat{k}\)
Since given vectors \(\vec{a}, \vec{b} \text { and } \vec{c}\) are mutually orthogonal.
∴ \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}\) = 0
Now \(\vec{a} \cdot \vec{c}\) = 0
⇒ \((\hat{i}-\hat{j}+2 \hat{k}) \cdot(\lambda \hat{i}+\hat{j}+\mu \hat{k})\) = 0
⇒ λ – 1 + 2µ = 0
⇒ λ + 2µ = 1 ………………………………(1)
and \(\vec{b} \cdot \vec{c}\) = 0
⇒ \((2 \hat{i}+4 \hat{j}+\hat{k}) \cdot(\lambda \hat{i}+\hat{j}+\mu \hat{k})\) = 0
⇒ 2λ + 4 + µ = 0
⇒ 2λ + µ = – 4 ………………..(2)
On solving (1) and (2) ; we have
λ = – 3 ; µ = 2

Question 33.
Find the values of λ and µ if the vectors \(\lambda \hat{i}-3 \hat{j}-6 \hat{k}\) and \(3 \hat{i}-\mu \hat{j}-2 \hat{k}\) are mutually perpendicular vectors of equal magnitude.
Solution:
Let \(\vec{a}=\lambda \hat{i}-3 \hat{j}-6 \hat{k}\)
and \(\vec{b}=3 \hat{i}-\mu \hat{j}-2 \hat{k}\)
Since \(\vec{a} \text { and } \vec{b}\) are mutually ⊥ to each other.
∴ \(\vec{a} \cdot \vec{b}\) = 0
⇒ \((\lambda \hat{i}-3 \hat{j}-6 \hat{k}) \cdot(3 \hat{i}-\mu \hat{j}-2 \hat{k})\) = 0
⇒ λ (3) – 3 (- µ) – 6 (- 2) = 0
⇒ 3λ + 3µ + 12 = 0
⇒ λ + µ + 4 = 0 ………………..(1)
also \(|\vec{a}|=|\vec{b}|\)
⇒ \(\sqrt{\lambda^2+(-3)^2+(-6)^2}=\sqrt{3^2+(-\mu)^2+(-2)^2}\)
⇒ \(\sqrt{\lambda^2+9+36}=\sqrt{9+\mu^2+4}\)
⇒ \(\sqrt{\lambda^2+45}=\sqrt{13+\mu^2}\)
On squaring ; we have
λ² – µ² = 13 – 45 = – 32
⇒ (λ – µ) (λ + µ) = – 32
⇒ (λ – µ) (- 4) = – 32 [using (1)]
⇒ λ – µ = 8 ……………………(2)
On adding (1) and (2) ; we have
2λ = 4
⇒ λ = 2
∴ from (1) ;
µ = – 6.

Question 34.
If \(\vec{a}=\hat{i}+\hat{j}-\hat{k}\) and \(\vec{b}=\hat{i}-\hat{j}+\hat{k}\), then find a unit vector which is perpendicular to \(\vec{a}\) and is coplanar with \(\vec{a} \text { and } \vec{b}\).
Solution:
Let \(\vec{r}\) be any vector in the plane of \(\vec{a} \text { and } \vec{b}\).
Then \(\vec{r}=\lambda \vec{a}+\mu \vec{b}\),
where λ, µ are scalars.

Question 35.
Find a vector of magnitude 5 units which is coplanar with vectors \(3 \hat{i}-\hat{j}-\hat{k}\) and \(\hat{i}+\hat{j}-2 \hat{k}\) and is perpendicular to the vector \(2 \hat{i}+2 \hat{j}+\hat{k}\).
Solution:
Let the required vector be \(\vec{d}=a \hat{i}+b \hat{j}+c \hat{k}\)
Since \(|\vec{d}|\) = 5 units
⇒ a² + b² + c² = 25
Let given vectors are \(\vec{\alpha}=3 \hat{i}-\hat{j}-\hat{k}\)
and \(\vec{\beta}=\hat{i}+\hat{j}-2 \hat{k}\)
Since \(\vec{d}\) is coplanar with \(\vec{\alpha} \text { and } \vec{\beta}\).

⇒ 2a + 2b + c = 0 ………………(5)
Using eqn. (2), (3) and (4) in eqn. (5) ; we have
2 (3λ + μ) + 2 (- λ + μ) + (- λ – 2μ) = 0
⇒ 3λ + 2μ = 0
⇒ μ = – \(\frac{3 \lambda}{2}\) ………………(6)
∴ eqn. (2) ; (3) and (4) becomes ;

Question 36.
(i) If \(\vec{a}\),\(\vec{b}\) and \(\vec{c}\) are vectors such that \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\), then show taht the angle θ between \(\vec{b} \text { and } \vec{c}\) is given that cos θ = \(=\frac{|\vec{a}|^2-|\vec{b}|^2-|\vec{c}|^2}{2|\vec{b}||\vec{c}|}\).
(ii) If \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\) and \(|\vec{a}|=5,|\vec{b}|=6 \quad \text { and } \quad|\vec{c}|=9\), then find the angle between \(\vec{a} \text { and } \vec{b}\).
Solution:
(i)

(ii) Given \(|\vec{a}|=5\) ;
\(|\vec{b}|=6\) ;
and \(|\vec{c}|=9\)
Let θ be the angle between \(\vec{a} \text { and } \vec{b}\)

Question 38.
In any ∆ABC, prove by vector method that cos B = \(\frac{c^2+a^2-b^2}{2 c a}\). (ISC 2010).
Solution:
Let \(\overrightarrow{\mathrm{BC}}, \overrightarrow{\mathrm{CA}}, \overrightarrow{\mathrm{AB}}\) represents the vectors \(\vec{a}, \vec{b} \text { and } \vec{c}\) respectively
Now \(\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}+\overrightarrow{\mathrm{AB}}=\overrightarrow{0}\)

Question 39.
Prove by vector method that a diameter of a circle will subtend a right angle at a point on its circumference. (ISC 2003)
Solution:
Let O be the centre of circle with AB as diameter
and P be any point on the circumference of circle.
Take O as origin
let \(\overrightarrow{\mathrm{OP}}=\vec{b}\) ;
\(\overrightarrow{\mathrm{OB}}=\vec{a}\) ;
∴ \(\overrightarrow{\mathrm{OA}}=-\vec{a}\)
Now we want to prove that
∠APB = 90°

∴ ∠APB = 90°
Hence, the diameter of a circle will subtend a right angle at a point on its circumference.

Question 40.
Prove by vector method that the diagonals of a rectangle are perpendicular if and only if the rectangle is a square.
Solution:
Let ABCD is a rectangle.

Students appreciate clear and concise ML Aggarwal Maths for Class 12 Solutions Chapter 1 Vectors Ex 1.1 that guide them through exercises.

ML Aggarwal Class 12 Maths Solutions Section B Chapter 1 Vectors Ex 1.1

Question 1.
Classify the following measures as scalars and vectors:
(i) 5 seconds
(ii) 20 m/sec²
(iii) 1000 cm³
(iv) 2 metres north-west
(v) 2 km. (NCERT)
Solution:
(i) 5 seconds is a time so it is a scalar, since it has only magnitude.
(ii) 20 rn/sec² is acceleration, since it has both
magnitude and direction hence it is a vector quantity.
(iii) 1000 cm³ is volume so it has only magnitude and hence represents a scalar quantity.
(iv) 2 metres north-west is displacement so it has both magnitude and direction and hence represents a vector quantity.
(v) 40 watt is voltage so it has only magnitude and hence represents a scalar.
(vi) 2 km is distance so it has only magnitude and hence represents a scalar.

Question 1 (old).
(i) Represent graphically a displacement of 30 km, 30° east of north. (NCERT)
(ii) Represent graphically a displacement of 40 km, 30° west of south. (NCERT)
Solution:
(i) Here \(\overrightarrow{\mathrm{OP}}\) represents the displacement of 30 km, 30° east of north.

Scale 1 cm = 20 km

(ii) Here \(\overrightarrow{\mathrm{OP}}\) represents a displacement of 40 km, 30° west of south

Scale 1 cm = 20 km

Question 2.
The adjoining figure is a square. Identify the following vectors :
(i) Equal
(ii) Co-initial
(iii) Collinear but not equal (NCERT)

Solution:
(i) Two vectors \(\vec{a} \text { and } \vec{b}\) are equal if they have equal magnitude, same or parallel support and same direction.
∴ \(\vec{b} \text { and } \vec{d}\) are equal vectors, since all sides of square are equal.
(ii) vectors having same initial point are said to be co-initial vectors.
\(\vec{a} \text { and } \vec{d}\) are co-initial vectors.
(iii) Two or more vectors are said to be collinear iff they have same or parallel supports, irrespective of their magnitude and directions.
Clearly \(\vec{a} \text { and } \vec{c}\) are collinear but not equal vectors
while \(\vec{b} \text { and } \vec{d}\) are collinear and equal vectors.

Question 3.
State whether the following are true or false:
(i) Vectors \(\vec{a} \text { and }-\vec{a}\) a are collinear. (NCERT)
(ii) Two collinear vectors are always equal in magnitude. (NCERT)
(iii) Two vectors having same magnitude are collinear. (NCERT)
(iv) Two collinear vectors having same magnitude are equal. (NCERT)
(v) Zero vector is unique.
(vi) Modulus of vectors \(3 \vec{a} \text { and }-3 \vec{a}\) is same.
(vii) Resultant of \(\vec{a} \text { and }-\vec{a}\) is a proper vector.
(viii) Two equal and co-initial vectors must coincide.
(ix) (m + n) (\((\vec{a}+\vec{b})\)) = \(m \vec{a}+m \vec{b}+n \vec{a}+n \vec{b}\)
(x) (m n) \((\vec{a}+\vec{b})=m \vec{a}+n \vec{b}\)
(xi) \(\vec{a}=-\vec{b}\)
⇒ \(|\vec{a}|=|\vec{b}|\)
(xii) \(\vec{a}=\vec{b} \Rightarrow|\vec{a}|=|\vec{b}|\)
(xiii) \(|\vec{a}|=|\vec{b}| \Rightarrow \vec{a}=\vec{b}\)
(xiv) If \(\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{B C}}=\overrightarrow{0}\), then points O and C coincide.
Solution:
(i) True, since both vectors lies on a same line

(ii) False, since two or more vectors are said to be collinear iff they have same or || supports irrespective of their magnitudes and directions.

(iii) False, Since both vectors may have different supports.

(iv) False, Since both vectors may have different directions.

(v) False, \(\overrightarrow{\mathrm{AA}}, \overrightarrow{\mathrm{BB}}\) etc. are all coinitial vectors.

(vi) True, \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{BA}}\) are of same length i.e. \(|\overrightarrow{\mathrm{AB}}|=|\overrightarrow{\mathrm{BA}}|\)

(vii) False, since the resultant of \(\vec{a} \text { and }-\vec{a}\) is a zero vector.

(viii) True, Since coinitial vectors and equal vectors are of same magnitude.

(ix) True

(x) False, (m n) \((\vec{a}+\vec{b})=m n \vec{a}+m n \vec{b}\)

(xi) True, \(\vec{a}=-\vec{b}\)
⇒ \(|\vec{a}|=|-\vec{b}|=|\vec{b}|\)

(xii) True, \(\vec{a}=\vec{b}\)
⇒ \(|\vec{a}|=|\vec{b}|\)

(xiii) False,
⇒ \(\vec{a}=3 \hat{i}+4 \hat{j}\) ;
\(\vec{b}=4 i+3 \hat{j}\)
\(|\vec{a}|=|\vec{b}|=\sqrt{3^2+4^2}\) = 5 but \(\vec{a} \neq \vec{b}\)

(xiv) True, \(\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{0}\)
⇒ \(\overrightarrow{\mathrm{OB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{0}\) [using ∆ law of vectors]
⇒ \(\overrightarrow{\mathrm{OC}}=\overrightarrow{0}\)
∴ \(\overrightarrow{\mathrm{O}} \text { and } \overrightarrow{\mathrm{C}}\) are coincide i.e. O and C are coincide.

Question 4.
If \(\vec{a} \text { and } \vec{b}\) are collinear, then state whether the following statements are true or false :
(i) b = λ . \(\vec{a}\) for some scalar.
(ii) \(\vec{a}= \pm \vec{b}\)
(iii) the respective components of \(\vec{a} \text { and } \vec{b}\) are proportional.
(iv) both the vectors \(\vec{a} \text { and } \vec{b}\) have same direction, but different magnitude.
Solution:
Since we know that, two or more vectors are collinear
iff they have same or parallel supports irrespective of their magnitudes and directions.

(i) True, since \(\vec{a} \text { and } \vec{b}\) are scalar multiple of each other.
∴ \(\vec{a}=\lambda \vec{b}\)
and \(\vec{b}=\mu \vec{a}\)
for some scalars λ and μ.

(ii) false, since \(\vec{a} \text { and } \vec{b}\) are collinear
∴ \(\vec{a}=\lambda \vec{b}\) for some scalar λ
∴ \(\vec{a} \neq \pm \vec{b}\) for λ ≠ ± 1

(iii) True, Let \(\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\)
and \(\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\)
Since \(\vec{a}=\lambda \vec{b}\) for some scalar λ.
∴ \(\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right)=\lambda\left(b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k}\right)\)
⇒ a₁ = λ b₁ ;
a₂ = λ b₂ ;
a₃ = λ b₃
Thus their components are proportional

(iv) False, Since \(\vec{a}=\lambda \vec{b}\) for some scalar λ.
Since \(\vec{a}=-2 \vec{b}\) for λ = – 2,
Here \(\vec{a} \text { and } \vec{b}\) are of opposite direction but \(\vec{a} \text { and } \vec{b}\) are collinear.

Question 5.
(i) If k \(\vec{a}=\overrightarrow{0}\), then what can you say about k and \(\vec{a}\) ?
(ii) If \(\vec{a}\) is a non-zero vector, then find a scalar k such that |k \(\vec{a}\)| = 1.
Solution:
(i) k \(\vec{a}\) = \(\overrightarrow{0}\)
⇒ k = 0 or \(\vec{a}=\overrightarrow{0}\)
or k and \(\vec{a}\) both are \(\overrightarrow{0}\).

(ii)

Question 5 (old).
In a triangle ABC (shown in the adjoining figure), state whether the following are true or false :
(i) \(\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)
(ii) \(\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{B C}}-\overrightarrow{\mathbf{A C}}=\overrightarrow{0}\)
(iii) \(\overrightarrow{\mathbf{A B}}-\overrightarrow{\mathbf{C B}}+\overrightarrow{\mathbf{C A}}=\overrightarrow{0}\)
(iv) \(\overrightarrow{\mathbf{A B}}+\overrightarrow{\mathbf{B C}}-\overrightarrow{\mathbf{C A}}=\overrightarrow{0}\)

Solution:
(i) By ∆ law of addition of vectors
\(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{AC}}\)
⇒ \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}-\overrightarrow{\mathrm{AC}}=\overrightarrow{0}\) ……………..(1)
⇒ \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)
∴ given statement is true.

(ii) From eqn. (1) ; given result is true.

(iii) True, Since \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)
⇒ \(\overrightarrow{\mathrm{AB}}-\overrightarrow{\mathrm{CB}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)
[∵ \(\overrightarrow{\mathrm{BC}}=-\overrightarrow{\mathrm{CB}}\)]

(iv) False, since \(\overrightarrow{\mathrm{AB}}+\overrightarrow{\mathrm{BC}}+\overrightarrow{\mathrm{CA}}=\overrightarrow{0}\)

Question 6.
If \(|\vec{a}|\) = 5, then calculate
(i) |2 \(\vec{a}\)|
(ii) \(\left|-\frac{1}{2} \vec{a}\right|\)
(iii) \(-\frac{1}{2}|\vec{a}|\)
Solution:
Given \(|\vec{a}|\) = 5
(i) \(|2 \vec{a}|\) = |2| \(|\vec{a}|\)
= 2 × 5 = 10

(ii) \(\left|-\frac{1}{2} \vec{a}\right|=\left|-\frac{1}{2}\right||\vec{a}|\)
= \(\frac{1}{2}\) × 5
= \(\frac{5}{2}\)

(iii) \(-\frac{1}{2}|\vec{a}|\)
= – \(\frac{1}{2}\) × 5
= – \(\frac{5}{2}\)

Question 7.
(i) For what value of p, is \((\hat{i}+\hat{j}+\hat{k})\) p a unit vector.
(ii) If \(\vec{a}=-3 \hat{i}-2 \hat{j}+6 \hat{k}\), then find the value of λ so that λ \(\vec{a}\) may be a unit vector.
Solution:
(i) Let \(\vec{a}\) = p \((\hat{i}+\hat{j}+\hat{k})\)
= \(p \hat{i}+p \hat{j}+p \hat{k}\)
Since \(\vec{a}\) be a unit vector.
∴ \(|\vec{a}|\) = 1
⇒ \(\sqrt{p^2+p^2+p^2}\) = 1
⇒ \(\sqrt{3 p^2}\) = 1
⇒ √3 p = 1
⇒ √3 |p| = 1
⇒ |p| = \(\frac{1}{\sqrt{3}}\)
⇒ p = ± \(\frac{1}{\sqrt{3}}\)

(ii) Given \(\vec{a}=-3 \hat{i}-2 \hat{j}+6 \hat{k}\)
Since λ \(\vec{a}\) is a unit vector

Question 8.
(i) Find values of x and y so that the vectors \(2 \hat{i}+3 \hat{j} \text { and } x \hat{i}+y \hat{j}\) are equal. (NCERT)
(ii) Find the values of x, y and z so that the vectors \(\vec{a}=x \hat{i}+2 \hat{j}+z \hat{k}\) and \(\vec{b}=2 \hat{i}+y \hat{j}+\hat{k}\) are equal. (NCERT)
(iii) If \(\vec{a}=x \hat{i}+2 \hat{j}-z \hat{k}\) and \(\vec{b}=3 \hat{i}-y \hat{j}+\hat{k}\) are two equal vectors, then write the value of x + y + z.
(iv) If \(\vec{a}=x \hat{i}+2 \hat{j}-3 \hat{k}\) and \(\vec{b}=3 \hat{i}-y \hat{j}+z \hat{k}\), then find the values of x, y and z so that \(2 \vec{a}=3 \vec{b}\).
Solution:
(i) Let \(\vec{a}=2 \hat{i}+3 \hat{j}\)
and \(\vec{b}=x \hat{i}+y \hat{j}\)
Since \(\vec{a} \text { and } \vec{b}\) are equal.
∴ \(\vec{a}=\vec{b}\)
⇒ \(2 \hat{i}+3 \hat{j}=x \hat{i}+y \hat{j}\)
⇒ x = 2 and y = 3.
[on comparing the coefficients of i and j on both sides]

(ii) Since \(\vec{a}=\vec{b}\)
⇒ \(x \hat{i}+2 \hat{j}+z \hat{k}=2 \hat{i}+y \hat{j}+\hat{k}\)
on comparing the coefficients of \(\hat{i}, \hat{j} \text { and } \hat{k}\) on both sides; we have
x = 2 ; y = 2 ; z = 1

(iii) Given \(\vec{a}=x \hat{i}+2 \hat{j}-z \hat{k}\)
and \(\vec{b}=3 \hat{i}-y \hat{j}+\hat{k}\)
Now \(\vec{a}=\vec{b}\)
⇒ \(x \hat{i}+2 \hat{j}-z \hat{k}=3 \hat{i}-y \hat{j}+\hat{k}\)
∴ x = 3 ; y = – 2 and z = – 1.
Thus x + y + z = 3 – 2 – 1 = 0

(iv) Given \(\vec{a}=x \hat{i}+2 \hat{j}-3 \hat{k}\)
and \(\vec{b}=3 \hat{i}-y \hat{j}+z \hat{k}\)
Since \(2 \vec{a}=3 \vec{b}\)
⇒ \(2(x \hat{i}+2 \hat{j}-3 \hat{k})=3(3 \hat{i}-y \hat{j}+z \hat{k})\)
⇒ \(2 x \hat{i}+4 \hat{j}-6 \hat{k}=9 \hat{i}-3 y \hat{j}+3 z \hat{k}\)
[since two vectors are equal iff their corresponding components are equal]
2x = 9 ;
4 = – 3y
and – 6 = 3z
⇒ x = \(\frac{9}{2}\) ;
y = – \(\frac{4}{3}\)
and z = – 2

Question 9.
(i) Let \(\vec{a}=\hat{i}+2 \hat{j}\) and \(\vec{b}=2 \hat{i}+\hat{j}\). Is \(|\vec{a}|=|\vec{b}|\) ? Are the vectors \(\vec{a} \text { and } \vec{b}\) equal ? (NCERT)
(ii) Write two different vectors having same magnitude. (NCERT)
(iii) Write two different vectors having same direction. (NCERT)
(iv) If \(|\vec{a}|=|\vec{b}|\), is it true that \(\vec{a}= \pm \vec{b}\) ? Justify your answer.
Solution:
(i) Given \(\vec{a}=\hat{i}+2 \hat{j}\)
and \(\vec{b}=2 \hat{i}+\hat{j}\)
Here \(|\vec{a}|=\sqrt{1^2+2^2}=\sqrt{5}\)
and \(|\vec{b}|=\sqrt{2^2+1^2}=\sqrt{5}\)
∴ \(|\vec{a}|=|\vec{b}|\)
Clearly \(\vec{a} \text { and } \vec{b}\) are not equal.

(ii) Let \(\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}\) ;
\(|\vec{a}|=\sqrt{2^2+(-1)^2+2^2}\) = 3
& \(\vec{b}=-2 \hat{i}+\hat{j}-2 \hat{k}\) ;
\(|\vec{b}|=\sqrt{(-2)^2+1^2+(-2)^2}=\sqrt{9}\) = 3

(iii) \(\vec{a}=2 \hat{i}+\hat{j}+2 \hat{k}\)
& \(\vec{b}=4 \hat{i}+2 \hat{j}+4 \hat{k}\)
Clearly \(\vec{b}=2 \vec{a}\)
i. e. \(\vec{b} \text { and } \vec{a}\) have same direction.
Since both \(\vec{a} \text { and } \vec{b}\) are parallel vectors.

(iv) Let \(\vec{a}=3 \hat{i}+4 \hat{j}\)
⇒ \(|\vec{a}|=\sqrt{3^2+4^2}\) = 5
and \(\vec{b}=4 \hat{i}+3 \hat{j}\)
⇒ \(|\vec{b}|=\sqrt{4^2+3^2}\) = 5
∴ \(|\vec{a}|=|\vec{b}|\)
but \(\vec{a} \neq \vec{b}\) and \(\vec{a} \neq-\vec{b}\)

Question 10.
If a unit vector \(\vec{a}\) makes angles \(\frac{\pi}{3}\) with \(\hat{i}, \frac{\pi}{4} \text { with } \hat{j}\) and an acute angle θ with \(\hat{k}\), then find the value of θ.
Solution:

Question 11.
Find the magnitude of the following vectors :
(i) \(\hat{i}+\hat{j}+\hat{k}\)
(ii) \(2 \hat{i}-3 \hat{j}-7 \hat{k}\)
(iii) \(\frac{1}{\sqrt{3}} \hat{i}-\frac{1}{\sqrt{3}} \hat{j}+\frac{1}{\sqrt{3}} \hat{k}\) (NCERT)
Solution:
(i) Let \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\)
∴ \(|\vec{a}|=\sqrt{1^2+1^2+1^2}\)
= \(\sqrt{1+1+1}=\sqrt{3}\)

(ii) Let \(\vec{a}=2 \hat{i}-3 \hat{j}-7 \hat{k}\)
∴ \(|\vec{a}|=|2 \hat{i}-3 \hat{j}-7 \hat{k}|\)
= \(\sqrt{2^2+(-3)^2+(-7)^2}\)
= \(\sqrt{4+9+49}=\sqrt{62}\)

(iii) Let \(\vec{a}=\frac{1}{\sqrt{3}} \hat{i}-\frac{1}{\sqrt{3}} \hat{j}+\frac{1}{\sqrt{3}} \hat{k}\)
∴ \(|\vec{a}|=\sqrt{\left(\frac{1}{\sqrt{3}}\right)^2+\left(-\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2}\)
= \(\sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}\)
= \(\sqrt{\frac{3}{3}}=\sqrt{1}\) = 1

Question 12.
(i) Find the vector with initial point P (2, 3, 0) and terminal point Q (- 1, – 2, – 4). Also find its magnitude. (NCERT)
(ii) Write the scalar components of the vector \(\overrightarrow{\mathrm{AB}}\) with initial point A (2, 1) and terminal point B (- 5, 7).
(iii) Find the terminal point of the vector PQ whose initial point is P (- 2, 5, 0) and components along coordinate axes are 2, – 3 and 7 respectively.
Solution:
(i)

(ii) P.V of A = \(2 \hat{i}+\hat{j}\)
P.V of B = \(-5 \hat{i}+7 \hat{j}\)
∴ \(\overrightarrow{\mathrm{AB}}\) = P.V. of B – P.V. of A
= \(-5 \hat{i}+7 \hat{j}-(2 \hat{i}+\hat{j})\)
⇒ \(\overrightarrow{\mathrm{AB}}=-7 \hat{i}+6 \hat{j}\)
∴ Scalar components of vector \(\overrightarrow{\mathrm{AB}}\) are – 7 and 6.

(iii) Given, P.V of initial point P = \(-2 \hat{i}+5 \hat{j}+0 \hat{k}\)
Let Q be the terminal point of \(\overrightarrow{\mathrm{PQ}}\) with P.V \(\vec{b}\).
Since \(\overrightarrow{\mathrm{PQ}}\) is having components along coordinate axes are 2, – 3 and 7.
Thus, \(\overrightarrow{\mathrm{PQ}}\)
⇒ P.V of Q – P.V of P = \(2 \hat{i}-3 \hat{j}+7 \hat{k}\)
⇒ \(\vec{b}-(-2 \hat{i}+5 \hat{j}+0 \hat{k})=2 \hat{i}-3 \hat{j}+7 \hat{k}\)
⇒ \(\vec{b}=(2 \hat{i}-3 \hat{j}+7 \hat{k})+(-2 \hat{i}+5 \hat{j}+0 \hat{k})\)
⇒ \(\vec{b}=0 \hat{i}+2 \hat{j}+7 \hat{k}\)
Hence the terminal point of \(\overrightarrow{\mathrm{PQ}}\) be Q(0, 2, 7).

Question 13.
(i) Find the sum of the vectors \(\vec{a}=\hat{i}-2 \hat{j}+\hat{k}\), \(\vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}\) and \(\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}\).
(ii) If \(\vec{a}, \vec{b}, \vec{c}\) have components (1, – 1), (2, – 2) and (2, 1) respectively, find \(2 \vec{a}-\vec{b}+3 \vec{c}\).
Solution:
(i) Given \(\vec{a}=\hat{i}-2 \hat{j}+\hat{k}\),
\(\vec{b}=-2 \hat{i}+4 \hat{j}+5 \hat{k}\)
and \(\vec{c}=\hat{i}-6 \hat{j}-7 \hat{k}\)
∴ \(\vec{a}+\vec{b}+\vec{c}\) = \((1-2+1) \hat{i}+(-2+4-6) \hat{j}+(1+5-7) \hat{k}\)
= \(0 \hat{i}-4 \hat{j}-\hat{k}\)

(ii) Since \(\vec{a}, \vec{b} \text { and } \vec{c}\) have components (1, – 1), (2, – 2) and (2, 1).

Question 14.
Find a unit vector in the direction of the vector \(\vec{a}=2 \hat{i}-3 \hat{j}\).
(ii) Find a unit vector in the direction of the vector \(\vec{a}=2 \hat{i}+\hat{j}+2 \hat{k}\).
(iii) Find a unit vector in the direction of the vector \(\vec{a}=3 \hat{i}-2 \hat{j}+6 \hat{k}\).
Solution:
(î) required unit vector in the direction of \(\vec{a}=2 \hat{i}-3 \hat{j}\)
= \(\hat{a}=\frac{\vec{a}}{|\vec{a}|}=\frac{2 \hat{i}-3 \hat{j}}{|2 \hat{i}-3 \hat{j}|}=\frac{2 \hat{i}-3 \hat{j}}{\sqrt{2^2+(-3)^2}}\)
= \(\frac{2}{\sqrt{13}} \hat{i}-\frac{3}{\sqrt{13}} \hat{j}\)

(ii) Given \(\vec{b}=2 \hat{i}+\hat{j}+2 \hat{k}\)
∴ \(|\vec{b}|=\sqrt{2^2+1^2+2^2}\) = 3
Thus required unit vector in the direction of \(\vec{b}\)
\(\hat{b}=\frac{\vec{b}}{|\vec{b}|}\)
= \(\frac{2 \hat{i}+\hat{j}+2 \hat{k}}{3}\)
= \(\frac{2}{3} \hat{i}+\frac{1}{3} \hat{j}+\frac{2}{3} \hat{k}\)

(iii) Given \(\vec{a}=3 \hat{i}-2 \hat{j}+6 \hat{k}\)
∴ \(|\vec{a}|=\sqrt{3^2+(-2)^2+6^2}=\sqrt{49}\) = 7
Thus required unit vector in the direction of \(\vec{a}\)
= \(\hat{a}=\frac{\vec{a}}{|\vec{a}|}\)
= \(\frac{3 \hat{i}-2 \hat{j}+6 \hat{k}}{7}\)
= \(\frac{3}{7} \hat{i}-\frac{2}{7} \hat{j}+\frac{6}{7} \hat{k}\)

Question 15.
(i) Find a unit vector in the direction of \(\) where P and Q have coordinates (5, 0, 8) and (3, 3, 2) respectively. (NCERT Exemplar)
(ii) Find the unit vector in the direction of AB where A and B are the points (1, 2, 3) and (4, 5, 6) respectively. (NCERT)
Solution:
(i) Here \(\overrightarrow{\mathrm{PQ}}\) = P.V of Q – P.V of P
= \((3 \hat{i}+3 \hat{j}+2 \hat{k})-(5 \hat{i}+0 \hat{j}+8 \hat{k})\)
= \(-2 \hat{i}+3 \hat{j}-6 \hat{k}\)
∴ required unit vector in the direction of \(\overrightarrow{\mathrm{PQ}}=\hat{\mathrm{PQ}}\)
= \(\frac{\overrightarrow{P Q}}{|\overrightarrow{P Q}|}\)
= \(\frac{-2 \hat{i}+3 \hat{j}-6 \hat{k}}{\sqrt{(-2)^2+3^2+(-6)^2}}\)
= \(\frac{-2 \hat{i}+3 \hat{j}-6 \hat{k}}{\sqrt{4+9+36}}\)
= \(-\frac{2}{7} \hat{i}+\frac{3}{7} \hat{j}-\frac{6}{7} \hat{k}\)

(ii) \(\overrightarrow{\mathrm{OA}}\) = P.V. of A
= \(\hat{i}+2 \hat{j}+3 \hat{k}\)
and \(\overrightarrow{\mathrm{OB}}\) = P.V. of B
= \((4 \hat{i}+5 \hat{j}+6 \hat{k})\)
∴ \(\overrightarrow{\mathrm{OB}}-\overrightarrow{\mathrm{OA}}\) = P.V. of B – P.V. of A
= \((4 \hat{i}+5 \hat{j}+6 \hat{k})-(\hat{i}+2 \hat{j}+3 \hat{k})\)
= \(3 \hat{i}+3 \hat{j}+3 \hat{k}\)
∴ \(\hat{\mathrm{AB}}=\frac{\overrightarrow{\mathrm{AB}}}{|\overrightarrow{\mathrm{AB}}|}=\frac{3(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3^2+3^2+3^2}}\)
= \(\frac{(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}}\)

Question 16.
(i) Write a unit vector in the direction of sum of vectors \(\vec{a}=2 \hat{i}+2 \hat{j}-5 \hat{k}\) and \(\vec{b}=2 \hat{i}+\hat{j}-7 \hat{k}\).
(ii) If \(\vec{a}=\hat{i}+\hat{j}+2 \hat{k}\) and \(\vec{b}=2 \hat{i}+\hat{j}-2 \hat{k}\), find the unit vector in the direction of \(2 \vec{a}-\vec{b}\).
Solution:
(i) Given \(\vec{a}=2 \hat{i}+2 \hat{j}-5 \hat{k}\)
and \(\vec{b}=2 \hat{i}+\hat{j}-7 \hat{k}\)

(ii) Given \(\vec{a}=\hat{i}+\hat{j}+2 \hat{k}\)
and \(\vec{b}=2 \hat{i}+\hat{j}-2 \hat{k}\)

Question 17.
(i) Find a unit vector parallel to te sum of vectors \(\vec{a}=2 \hat{i}+4 \hat{j}-5 \hat{k}\) and \(\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}\). (NCERT)
(ii) If \(\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}\) and \(\vec{b}=2 \hat{i}+4 \hat{j}+9 \hat{k}\), then find a unit vector parallel to \(\vec{a}+\vec{b}\).
Solution:
(i) Given \(\vec{a}=2 \hat{i}+4 \hat{j}-5 \hat{k}\)
and \(\vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}\)

(ii) Given \(\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k}\)
and \(\vec{b}=2 \hat{i}+4 \hat{j}+9 \hat{k}\)

Question 18.
(i) Find a vector in the direction of the vector \(\vec{a}=\hat{i}-2 \hat{j}\), whose magnitude is 7 units.
(ii) Write a vector of magnitude 9 units in the direction of the vector \(-2 \hat{i}+\hat{j}+2 \hat{k}\).
(iii) Find vectors of magnitude 5 units which are parallel to the sum of vectors \(\vec{a}=2 \hat{i}+3 \hat{j}-\hat{k}\) and \(\vec{b}=\hat{i}-2 \hat{j}+\hat{k}\).
Solution:
(i) Given \(\vec{a}=\hat{i}-2 \hat{j}\)
∴ required vector in the direction of \(\vec{a}\) having magnitude 7 units.
= \(7 \hat{a}=\frac{7 \vec{a}}{|\vec{a}|}\)
= \(\frac{7(\hat{i}-2 \hat{j})}{\sqrt{1^2+(-2)^2}}\)
= \(\frac{7}{\sqrt{5}}(\hat{i}-2 \hat{j})\)

(ii) We required a vector of magnitude 9 which is in the direction of \(\vec{a}=-2 \hat{i}+\hat{j}+2 \hat{k}\)
i.e. required vector = \(\frac{9 \vec{a}}{|\vec{a}|}\)
= \(\frac{9(-2 \hat{i}+\hat{j}+2 \hat{k})}{\sqrt{(-2)^2+1^2+2^2}}\)
= \(3(-2 \hat{i}+\hat{j}+2 \hat{k})\)

(iii) Given \(\vec{a}=2 \hat{i}+3 \hat{j}-\hat{k}\) ;
and \(\vec{b}=\hat{i}-2 \hat{j}+\hat{k}\)
∴ \(\vec{a}+\vec{b}=3 \hat{i}+\hat{j}+0 \hat{k}\)
\(|\vec{a}+\vec{b}|=\sqrt{3^2+1^2+0^2}=\sqrt{10}\)
Thus, required unit vector of magnitude 5
= \(\frac{ \pm 5(\vec{a}+\vec{b})}{|(\vec{a}+\vec{b})|}\)
= \(\frac{ \pm 5(3 \hat{i}+\hat{j})}{\sqrt{10}}\)

Question 18 (old).
(i) If \(\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}\) and \(\vec{b}=-\hat{i}+\hat{j}-\hat{k}\), then find the unit vector in the direction of the vector \(\vec{a}+\vec{b}\). (NCERT)
(ii) If \(\vec{a}=\hat{i}+\hat{j}\), \(\vec{b}=\hat{j}+\hat{k}\) and \(\vec{c}=\hat{k}+\hat{i}\) then find a unit vector in the direction of \(\vec{a}+\vec{b}+\vec{c}\).
Solution:
(i) Given \(\vec{a}=2 \hat{i}-\hat{j}+2 \hat{k}\)
and \(\vec{b}=-\hat{i}+\hat{j}-\hat{k}\)
∴ \(\vec{a}+\vec{b}=(2 \hat{i}-\hat{j}+2 \hat{k})+(-\hat{i}+\hat{j}-\hat{k})\)
= \(\hat{i}+0 \hat{j}+\hat{k}\)
⇒ \(\vec{c}=\vec{a}+\vec{b}\)
= \(\hat{i}+0 \hat{j}+\hat{k}\)
Thus unit vector in the direction \(\vec{a}+\vec{b}\)
∴ \(\hat{c}=\frac{\vec{c}}{|\vec{c}|}\)
= \(\frac{\hat{i}+0 \hat{j}+\hat{k}}{\sqrt{1^2+1^2}}\)
= \(\frac{1}{\sqrt{2}}(\hat{i}+\hat{k})\)

(iv) Given \(\vec{a}=\hat{i}+\hat{j}\) ;
\(\vec{b}=\hat{j}+\hat{k}\) ;
and \(\vec{c}=\hat{k}+\hat{i}\)

Question 19.
(i) If the magnitude of the position vector of the point (3, – 2, α) is 7 units, then find all possible values of α.
(ii) What vector should be added to the vector \(\vec{a}=3 \hat{i}+4 \hat{j}-2 \hat{k}\) so that the resultant is a unit vector \(\hat{i}\)?
Solution:
(i) Let \(\vec{a}\) be the position vector of point (3, – 2, α)
∴ \(\vec{a}=3 \hat{i}-2 \hat{j}+\alpha \hat{k}\) also given \(|\vec{a}|\) = 7
⇒ \(\sqrt{3^2+(-2)^2+\alpha^2}\) = 7
⇒ \(\sqrt{9+4+\alpha^2}\) = 7
⇒ \(\sqrt{13+\alpha^2}\) = 7
Oh squaring; we have
13 + α² = 49
⇒ α² = 49 – 13 = 36
⇒ α = ± 6

(ii) Given \(\vec{a}=3 \hat{i}+4 \hat{j}-2 \hat{k}\)
Let the required vector be \(\vec{b}\) so that
\(\vec{a}+\vec{b}=\hat{i}\)
⇒ \(\vec{b}=\hat{i}-\vec{a}\)
= \(\hat{i}-(3 \hat{i}+4 \hat{j}-2 \hat{k})\)
⇒ \(\vec{b}=-2 \hat{i}-4 \hat{j}+2 \hat{k}\)

Question 20.
(i) Show that the vectors \(2 \hat{i}-3 \hat{j}+4 \hat{k}\) and \(-4 \hat{i}+6 \hat{j}-8 \hat{k}\) are collinear.
(ii) Write the value of p for which \(\vec{a}=3 \hat{i}+2 \hat{j}+9 \hat{k}\) and \(\vec{b}=\hat{i}+p \hat{j}+3 \hat{k}\) are parallel vectors.
(iii) If \(\vec{a}=p \hat{i}+3 \hat{j}\) and \(\vec{b}=4 \hat{i}+p \hat{j}\), then find the values of p so that \(\vec{a} \text { and } \vec{b}\) may be parallel.
Solution:
(i) Let O be the origin and A, B be the points whose position vectors are

Thus \(\overrightarrow{\mathrm{OA}} \text { and } \overrightarrow{\mathrm{OB}}\) vectors are parallel vectors and the origin O is common to them.
Thus \(\overrightarrow{\mathrm{OA}} \& \overrightarrow{\mathrm{OB}}\) vectors are collinear,
Therefore A, B are collinear vectors.

(ii) Given \(\vec{a}=3 \hat{i}+2 \hat{j}+9 \hat{k}\)
\(\vec{b}=\hat{i}+p \hat{j}+3 \hat{k}\) are parallel vectors
∴ \(\vec{a}=\lambda \vec{b}\) for some non-zero scalar λ.
⇒ \(3 \hat{i}+2 \hat{j}+9 \hat{k}=\lambda(\hat{i}+p \hat{j}+3 \hat{k})\)
∴ 3 = λ ;
2 = λ p
and 9 = 3λ
Thus, 2 = 3p
⇒ p = \(\frac{2}{3}\)

(iii) Given \(\vec{a}=p \hat{i}+3 \hat{j}\)
and \(\vec{b}=4 \hat{i}+p \hat{j}\)

Question 20 (old).
(iii) Find the vector in the direction of the vector \(\vec{r}=5 \hat{i}-\hat{j}+2 \hat{k}\) which has magnitude 8 units.
(iv) Find vectors of magnitude 5 units which are parallel to the vector \(3 \hat{i}+\hat{j}\).
Solution:
(iii) Let \(\vec{a}=5 \hat{i}-\hat{j}+2 \hat{k}\),
Let < l, m, n > be the direction cosines of \(\vec{a}\)
The direction ratios of \(\vec{a}\) are proportional to < 5, – 1, 2 >
∴ direction cosines of \(\vec{a}\) are \(\left\langle\frac{5}{\sqrt{5^2+(-1)^2+2^2}}, \frac{-1}{\sqrt{5^2+(-1)^2+2^2}}, \frac{2}{\sqrt{5^2+(-1)^2+2^2}}\right\rangle\)
i.e. \(\left\langle\frac{5}{\sqrt{30}}, \frac{-1}{\sqrt{30}}, \frac{2}{\sqrt{30}}\right\rangle\)
required vector = 8 \((\ell \hat{i}+m \hat{j}+n \hat{k})\)
= 8 \(\left(\frac{5}{\sqrt{30}} \hat{i}-\frac{1}{\sqrt{30}} \hat{j}+\frac{2}{\sqrt{30}} \hat{k}\right)\)
= \(\frac{8}{\sqrt{30}}(5 \hat{i}-\hat{j}+2 \hat{k})\)

(iv) Given \(\vec{a}=3 \hat{i}+\hat{j}\)
So a unit vector parallel to \(\vec{a}= \pm \hat{a}\)
= \(\pm \frac{\vec{a}}{|\vec{a}|}\)
= \(\pm \frac{(3 \hat{i}+\hat{j})}{\sqrt{3^2+1}}\)
= \(\pm \frac{(3 \hat{i}+\hat{j})}{\sqrt{10}}\)
required vector = \(5 \hat{a}= \pm \frac{5}{\sqrt{10}}(3 \hat{i}+\hat{j})\)

Question 21.
(i) Write the direction ratios of the vector \(\hat{i}+\hat{j}-2 \hat{k}\), and hence calculate its direction cosines.
(ii) Find the direction cosines of the vector joining the points A (1, 2, – 3) and B (- 1, 2, 1), directed from A to B.
Solution:
(i) Let \(\vec{a}=\hat{i}+\hat{j}-2 \hat{k}\)
∴ D ratios of \(\vec{a}\) be < 1, 1, – 2 >
Thus, D’ cosines of \(\vec{a}\) are
< \(\frac{1}{\sqrt{1^2+1^2+(-2)^2}}, \frac{1}{\sqrt{1^2+1^2+(-2)^2}}, \frac{-2}{\sqrt{1^2+1^2+(-2)^2}}\) >
i.e., < \(\frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{-2}{\sqrt{6}}\) >

(ii) direction ratios of vector \(\overrightarrow{\mathrm{AB}}\) are ;
< – 1 – 1 , – 2 – 2, 1 + 3 >
i.e., < – 2, – 4, 4 >
i.e., < – 1 , – 2, 2 >
Thus direction cosines of vector \(\overrightarrow{\mathrm{AB}}\) are ;
< \(\frac{-1}{\sqrt{(-1)^2+(-2)^2+2^2}}, \frac{-2}{\sqrt{(-1)^2+(-2)^2+2^2}}, \frac{2}{\sqrt{(-1)^2+(-2)^2+2^2}}\) >
i.e., < \(-\frac{1}{3},-\frac{2}{3}, \frac{2}{3}\) >

Question 22.
(i) Find the mid – point of the vector joining the points P and Q having position vectors \(2 \hat{i}+3 \hat{j}-4 \hat{k}\) and \(4 \hat{i}+\hat{j}-2 \hat{k}\) respectively.
(ii) The position vector of the mid-point M of the line segment AB is \(3 \hat{i}+4 \hat{k}\). If the position vector of the point A is \(4 \hat{i}-2 \hat{j}+3 \hat{k}\), find the position vector of B.
Solution:
(i) Given P.V. of P = \(2 \hat{i}+3 \hat{j}-4 \hat{k}\)
and P.V. of Q = \(4 \hat{i}+\hat{j}-2 \hat{k}\)

∴ P.V. of mid-point of line joining P and Q = \(\frac{(2 \hat{i}+3 \hat{j}-4 \hat{k})+(4 \hat{i}+\hat{j}-2 \hat{k})}{2}\)
= \(\frac{6 \hat{i}+4 \hat{j}-6 \hat{k}}{2}\)
= \(3 \hat{i}+2 \hat{j}-3 \hat{k}\)

(ii) Let P.V. of B = \(\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}\)
given P.V. of A = \(4 \hat{i}-2 \hat{j}+3 \hat{k}\)

Question 22 (old).
(ii) Write the position vector of the point dividing the line segment joining the points with position vector \(\vec{a} \text { and } \vec{b}\) externally in the ratio 1 : 4, where \(\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}\) and \(\vec{b}=-\hat{i}+\hat{j}+\hat{k}\)
Solution:
Given \(\vec{a}=2 \hat{i}+3 \hat{j}+4 \hat{k}\)
and \(\vec{b}=-\hat{i}+\hat{j}+\hat{k}\)
∴ \(\vec{a}=\lambda \vec{b}\)
⇒ \(3 \hat{i}+2 \hat{j}+9 \hat{k}=\lambda(\hat{i}+p \hat{j}+3 \hat{k})\)
∴ 3 = λ ;
2 = λ p
and 9 = 3λ
Thus, 2 = 3p
⇒ p = \(\frac{2}{3}\)

Question 23 (old).
(ii) Find the direction cosines of the vector \(\hat{i}+2 \hat{j}+3 \hat{k}\). (NCERT)
(iii) If P (1, 5, 4) and Q (4, 1, – 2), then find the direction ratios of \(\overrightarrow{\mathrm{PQ}}\).
Solution:
(ii) Let \(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\)
∴ direction ratio of are proportional to < 1, 2, 3 >
Thus, direction cosines of \(\vec{a}\) are ;
\(\left\langle\frac{1}{\sqrt{1^2+2^2+3^2}}, \frac{2}{\sqrt{1^2+2^2+3^2}}, \frac{3}{\sqrt{1^2+2^2+3^2}}\right\rangle\)
i.e., \(\left\langle\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}\right\rangle\)

(iii) We know that,
If P (x₁, y₁, z₁) and Q (x₂, y₂, z₂)
Then D’ ratios of \(\overrightarrow{\mathrm{PQ}}\) are
< x₂ – x₁, y₂ – y₁, z₂ – z₁ >
∴ D’ ratios of \(\overrightarrow{\mathrm{PQ}}\) are < 4 – 1, 1 – 5, – 2 – 4 >
i.e., < 3, – 4, – 6 >
Thus, D’ cosines of \(\overrightarrow{\mathrm{PQ}}\) are
< \(\frac{3}{\sqrt{3^2+(-4)^2+(-6)^2}}, \frac{-4}{\sqrt{3^2+(-4)^2+(-6)^2}}, \frac{-6}{\sqrt{3^2+(-4)^2+(-6)^2}}\) >
i.e., < \(\frac{3}{\sqrt{9+16+36}}, \frac{-4}{\sqrt{9+16+36}}, \frac{-6}{\sqrt{9+16+36}}\) >
i.e., < \(\frac{3}{\sqrt{61}}, \frac{-4}{61}, \frac{-6}{61}\) >

Question 23.
If the position vectors of theverticesA, B and C of a ∆ABC are \(\hat{i}+2 \hat{j}-3 \hat{k}\), \(-2 \hat{i}-5 \hat{j}+4 \hat{k}\) and \(7 \hat{i}-4 \hat{k}\) respectively, then write the position vector of the centrold of ∆ABC.
Solution:
Let \(\vec{a}, \vec{b}, \vec{c}\) be the position vectors of the vertices A, B and C of ∆ABC.
Then, \(\vec{a}=\hat{i}+2 \hat{j}-3 \hat{k\)k ;
\(\vec{b}=-2 \hat{i}-5 \hat{j}+4 \hat{k\)
and \(\vec{c}=7 \hat{i}-4 \hat{k}\)
Thus the P.V of centroid of ∆ABC are \(\left(\frac{\vec{a}+\vec{b}+\vec{c}}{3}\right)\)
= \(\frac{(\hat{i}+2 \hat{j}-3 \hat{k})+(-2 \hat{i}-5 \hat{j}+4 \hat{k})+(7 \hat{i}-4 \hat{k})}{3}\)
= \(\frac{6 \hat{i}-3 \hat{j}-3 \hat{k}}{3}\)
= \(2 \hat{i}-\hat{j}-\hat{k}\)

Question 24.
Find the position vector of a point R which divides the line segment joining the points P and Q with position vectors \(\) and \(\) respectively in the ratio 2 : 1
(i) internally
(ii) externally. (NCERT)
Solution:
(i) P.V. of point R
= \(\frac{2(-\hat{i}+\hat{j}+\hat{k})+1(\hat{i}+2 \hat{j}-\hat{k})}{2+1}\)
= \(\frac{(-2+1) \hat{i}+(2+2) \hat{j}+(2-1) \hat{k}}{2+1}\)
= \(-\frac{\hat{i}}{3}+\frac{4}{3} \hat{j}+\frac{1}{3} \hat{k}\)

(ii) P.V. of poiont R = \(\frac{2(-\hat{i}+\hat{j}+\hat{k})-1(\hat{i}+2 \hat{j}-\hat{k})}{2-1}\)
= \((-2-1) \hat{i}+(2-2) \hat{j}+(2+1) \hat{k}\)
= \(-3 \hat{i}+3 \hat{k}\)

Question 25.
(i) Find the position vector of the point which divides the join of the points with position vectors \(\vec{a}+3 \vec{b}\) and \(\vec{a}-\vec{b}\) internally in the ratio 1 : 3.
(ii) X and Y are two points with position vectors \(3 \vec{a}+\vec{b}\) and \(\vec{a}-3 \vec{b}\) respectively. Write the position vector of a point Z which divides the line segment XY in the ratio 2 : 1 externally.
Solution:
(i) Then by using section formula, P.V. of point which divides the line segment AB internally in the ratio 1 : 3

(ii) Thus by section formula.
The P.V of point which divides the line segment AB externally in the ratio 2: 1 i.e. 2 : – 1 internally

Question 26.
The vectors a and b are non-zero non-collinear. For what value of x, the vectors (x – 2) \(\) and (2x + 1) \(\) are collinear?
Solution:
Let \(\vec{c}=(x-2) \vec{a}+\vec{b}\)
and \(\vec{d}=(2 x+1) \vec{a}-\vec{b}\)
Since \(\vec{c}=\lambda \vec{d}\) are collinear.
∴ \(\vec{c} \text { and } \vec{d}\)
for some non-zero scalar λ.
⇒ (x – 2) \(\vec{a}+\vec{b}\) = λ [(2x + 1) \(\vec{a}-\vec{b}\)]
Since \(\vec{a} \text { and } \vec{b}\) are non-zero non-collinear vectors.
∴ x – 2 = λ (2x + 1).
and 1 = – λ
⇒ λ = – 1
∴ from (1) ;
x – 2 = λ (2x + 1)
⇒ x – 2 = – 2x – 1
⇒ 3x = 1
⇒ x = \(\frac{1}{3}\)

Question 27.
The position vectors of four points A, B, C, D are \(\vec{a}, \vec{b}, 2 \vec{a}+3 \vec{b}, 2 \vec{a}-3 \vec{b}\) respectively. Express the vectors \(\overrightarrow{\mathbf{A C}}, \overrightarrow{\mathbf{B D}}, \overrightarrow{\mathbf{C D}}, \overrightarrow{\mathbf{A B}}\) in terms of \(\vec{a} \text { and } \vec{b}\).
Solution:

Question 27 (old) .
(i) A and B are two points with position vectors \(2 \vec{a}-3 \vec{b}\) and \(6 \vec{b}-\vec{a}\) respectively. Write the position vector of a point P which divides the line segment AB internally in the ratio 1 : 2.
(ii) P and Q are points with position vectors \(3 \vec{a}-2 \vec{b}\) and \(\vec{a}+\vec{b}\) respectively. Write the position vector of a point R which divides the line segment PQ in the ratio 2: 1 externally.
Solution:
(i) required P.V of a point P which divides AB internally in the ratio 1 : 2 is given by

(ii) The P.V R of point which divides the line segment externally in the ratio 2 : 1.
i.e. internally 2 : – 1
\(\overrightarrow{\mathrm{R}}=\frac{2(\vec{a}+\vec{b})-(3 \vec{a}-2 \vec{b})}{2-1}\)
= \(-\vec{a}+4 \vec{b}\)

Question 28.
(i) If the points A, B, C and D have coordinates (0, 1), (1, 0), (1, 2) and (2, 1) respectively, then prove that \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{CD}}\).
(ii) If the coordinates of the points A and B in a plane are (1, 1) and (2, 2) respectively, then find the coordinates of point C such that \(\overrightarrow{\mathbf{A B}}=\overrightarrow{\mathrm{BC}}\).
Solution:
(i) Let \(\vec{a}, \vec{b}, \vec{c} \text { and } \vec{d}\) are the position vectors of points A, B, C and D respectively.

(ii) Given P.V of A = \(\hat{i}+\hat{j}\)
P.V of B = \(2 \hat{i}+2 \hat{j}\)
Let the coordinates of C be (a, b)
∴ P.V. of C = \(a \hat{i}+b \hat{j}\)
Since \(\overrightarrow{\mathrm{AB}}=\overrightarrow{\mathrm{BC}}\)
⇒ P.V of B – P.V of A = P.V of C – P.V of B
⇒ \((2 \hat{i}+2 \hat{j})-(\hat{i}+\hat{j})=(a \hat{i}+b \hat{j})-(2 \hat{i}+2 \hat{j})\)
⇒ \(\hat{i}+\hat{j}=(a-2) \hat{i}+(b-2) \hat{j}\)

Since if two vectors are equal then their corresponding components are also equal.
∴ 1 = a – 2
⇒ a = 3
and 1 = b – 2
⇒ b = 3.
Hence the coordinates of point C be (3, 3).

Question 29.
If the components of \(\vec{a}\) are (1, 2) and \(\vec{b}=\hat{i}-\hat{j}+\vec{a}\), then what are the componcnts of \(2 \vec{a}-3 \vec{b}\) ?
Solution:
Since the components of \(\vec{a}\) are (1, 2)
∴ \(\vec{a}=\hat{i}+2 \hat{j}\)
and \(\vec{b}=\hat{i}-\hat{j}+\vec{a}\)
= \(\hat{i}-\hat{j}+\hat{i}+2 \hat{j}\)
= \(2 \hat{i}+\hat{j}\)
∴ \(2 \vec{a}-3 \vec{b}=2(\hat{i}+2 \hat{j})-3(2 \hat{i}+\hat{j})\)
= \(2 \hat{i}+4 \hat{j}-6 \hat{i}-3 \hat{j}\)
= \(-4 \hat{i}+\hat{j}\)
∴ Components of \(2 \vec{a}-3 \vec{b}\) are (- 4, 1).

Question 30.
Let \(\overrightarrow{\mathrm{PQ}}\) be a vector of magnitude 8 units making an angle of 30° with x-axis lying in fourth quadrant. Find its components along x-axis and y-axis.
Solution:
Given |\(\overrightarrow{\mathrm{PQ}}\)| = 8 units
and \(\overrightarrow{\mathrm{PQ}}\) making an angle of 30° with x-axis and lying in fourth quadrant.
Let < l, m > be the direction cosine of \(\overrightarrow{\mathrm{PQ}}\)
∴ l = cos 330°
= cos (360° – 30°)
= cos 30°
= \(\frac{\sqrt{3}}{2}\)
and l² + m² = 1
m² = 1 – l²
m = – \(\sqrt{\left(1-\frac{3}{4}\right)}\)
= \(-\sqrt{\frac{1}{4}}=-\frac{1}{2}\)

or m = cos (240°)
= cos (180° + 60°)
= – cos 60°
= – \(\frac{1}{2}\)
∴ |\(\overrightarrow{\mathrm{PQ}}=|\overrightarrow{\mathrm{PQ}}|(l \hat{i}+m \hat{j})\)|
= 8 \(\left(\frac{\sqrt{3}}{2} \hat{i}-\frac{1}{2} \hat{j}\right)\)
⇒ \(\overrightarrow{\mathrm{PQ}}=4 \sqrt{3} \hat{i}-4 \hat{j}\)
Thus, components of \(\overrightarrow{\mathrm{PQ}}\) along x – axis and y – axis are ; 4√3 and – 4.

Question 31.
(i) Show that the points (0, – 1), (1, 0) and (3, 2) are collinear.
(ii) The position vectors of the points A, B, C are \(2 \hat{i}+\hat{j}-\hat{k}, 3 \hat{i}-2 \hat{j}+\hat{k}\) and \(\hat{i}+4 \hat{j}-3 \hat{k}\) respectively, show that A, B, C are collinear.
Solution:
(i) The position vectors of points A (0, – 1), B (1, 0) and C (3, 2) are \(-\hat{j}, \hat{i}, 3 \hat{i}+2 \hat{j}\).
∴ \(\overrightarrow{\mathrm{AB}}\) = P.V. of B – P.V. of A
= \(\hat{i}-(-\hat{j})\)
= \(\hat{i}+\hat{j}\)

∴ \(\overrightarrow{\mathrm{AC}}\) = P.V. of C – P.V. of A
= \(3 \hat{i}+2 \hat{j}-(-\hat{j})\)
= \(3 \hat{i}+3 \hat{j}\)

Since \(\overrightarrow{\mathrm{AB}}=\frac{1}{3} \overrightarrow{\mathrm{AC}}\)

Thus the vectors \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{AC}}\) are collinear.

∴ \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{AC}}\) are having same or parallel supports.

But \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{AC}}\) are coinitial vectors

∴ they are having same support. Thus, the points A, B and C are collinear.

(ii) Let A, B and Care the points whose position vectors are \(2 \hat{i}+\hat{j}-\hat{k}\), \(3 \hat{i}-2 \hat{j}+\hat{k}\) and \(\hat{i}+4 \hat{j}-3 \hat{k}\)

Thus \(\overrightarrow{\mathrm{BC}} \text { and } \overrightarrow{\mathrm{AB}}\) are parallel vectors and the point B is common to both vectors.
Thus \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{BC}}\) are collinear vectors and hence the points A, B, and c are collinear points.

Question 32.
If \(\vec{a}, \vec{b}\) are position vectors of points A (1, – 1) and B (- 3, m), then find the value of m so that the origin O, and points A and B are
(i) collinear
(ii) not collinear.
Solution:
(i) Given \(\vec{a}\) = P.V.of A
= \(\hat{i}-\hat{j}\)
and \(\vec{b}\) = P.V.of B
= \(-3 \hat{i}+m \hat{j}\)
Since the points O, A and B are collinear.
Thus \(\overrightarrow{\mathrm{OA}}, \overrightarrow{\mathrm{OB}}\) are collinear.
∴ \(\overrightarrow{\mathrm{OA}}=\lambda \overrightarrow{\mathrm{OB}}\) for sorne non-zero scalar λ.
⇒ \(\vec{a}-\overrightarrow{0}=\lambda(\vec{b}-\overrightarrow{0})\)
⇒ \(\hat{i}-\hat{j}=\lambda(-3 \hat{i}+m \hat{j})\)
∴ 1 = – 3 λ
and – 1 = λ m
⇒ λ = – \(\frac{1}{3}\)
and – 1 = – \(\frac{1}{3}\) m
m = 3.

(ii) Now points O, A, B are non collinear if m ≠ 3.

Question 33.
(i) if the points (α, – 1), (2, 1) and (4, 5) are collinear, then find a by vector method.
(ii) Using vectors, find he value of λ such that (λ, – 10, 3), (1, – 1, 3) amd (3, 5, 3) are collinear. (NCERT Exemplar)
Solution:
(i) Let the given points are A (α, – 1), B (2, 1) and C (4, 5)
∴ P.V of A = \(\alpha \hat{i}-\hat{j}\) ;
P.V of B = \(2 \hat{i}+\hat{j}\)
and P.V of C = \(4 \hat{i}+5 \hat{j}\)

∴ \(\overrightarrow{\mathrm{AB}}\) = P.V of B – P.V of A
= \((2 \hat{i}+\hat{j})-(\alpha \hat{i}-\hat{j})\)
= (2 – α) \(\hat{i}+2 \hat{j}\)

\(\overrightarrow{\mathrm{AC}}\) = P.V of C – P.V of A
= \((4 \hat{i}+5 \hat{j})-(\hat{i}-\hat{j})\)
= (4 – α) \(\hat{i}+6 \hat{j}\)

Since A, B and C are collinear points.
Thus \(\overrightarrow{\mathrm{AB}} \text { and } \overrightarrow{\mathrm{AC}}\) are collinear vectors.
Thus, \(\overrightarrow{\mathrm{AB}}=\lambda \overrightarrow{\mathrm{AC}}\) for some non-zero scalar λ.

⇒ \((2-\alpha) \hat{i}+2 \hat{j}\) = λ \([(4-\alpha) \hat{i}+6 \hat{j}]\)
∴ 2 – α = λ (4 – α)
and 2 = 6λ
λ = \(\frac{1}{3}\)
⇒ 2 – α = \(\frac{1}{3}\) (4 – α)
⇒ 6 – 3α = 4 – α
⇒ 3α – α = 6 – 4
⇒ 2α = 2
⇒ α = 1

(ii) Let A (λ, – 10, 3), B (1, – 1, 3) and C (3, 5, 3) are given points.
∴ \(\overrightarrow{\mathrm{AB}}\) = P.V.of B – P.V.of A
= \((\hat{i}-\hat{j}+3 \hat{k})-(\lambda \hat{i}-10 \hat{j}+3 \hat{k})\)
= \((1-\lambda) \hat{i}+9 \hat{j}+0 \hat{k}\)

& \(\overrightarrow{\mathrm{BC}}\) = P.V.of C – P.V.of B
= \((3 \hat{i}+5 \hat{j}+3 \hat{k})-(\hat{i}-\hat{j}+3 \hat{k})\)
= \(2 \hat{i}+6 \hat{j}+0 \hat{k}\)
Since A, B, C are collinear points.
∴ \(\overrightarrow{\mathrm{AB}}\) and \(\overrightarrow{\mathrm{BC}}\) are collinear vectors.
∴ \(\overrightarrow{\mathrm{AB}}=m \overrightarrow{\mathrm{BC}}\) for some scalar m.
⇒ \((1-\lambda) \hat{i}+9 \hat{j}=m[2 \hat{i}+6 \hat{j}]\)
Comparing the coefficients of \(\hat{i} \& \hat{j}\) on both sides, we have,
1 – λ = 2m ………………….(1)
and 9 = 6m
⇒ m = 3/2
∴ From (1) ; we have
1 – λ = 2 × \(\frac{3}{2}\)
⇒ 1 – λ = 3
⇒ λ = – 2

Question 33. (old).
(ii) Show that the points A (2, 6, 3), B (1, 2, 7) and C (3, 10, – 1) are collinear.
Solution:
Let O be the origin of reference

∴ \(\overrightarrow{\mathrm{AC}} \text { and } \overrightarrow{\mathrm{AB}}\) are parallel and also coinitial vectors.
Hence A, B, C are collinear.

Question 34.
If \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\), \(\vec{b}=4 \hat{i}-2 \hat{j}+3 \hat{k}\) and \(\vec{c}=\hat{i}-2 \hat{j}+\hat{k}\) then find a vector of magnitude 6 units which is parallel to the vector \(2 \vec{a}-\vec{b}+3 \vec{c}\).
Solution:
Given \(\vec{a}=\hat{i}+\hat{j}+\hat{k}\),
\(\vec{b}=4 \hat{i}-2 \hat{j}+3 \hat{k}\)
and \(\vec{c}=\hat{i}-2 \hat{j}+\hat{k}\)

Question 35.
(i) Show that the points A, B and C with position vectors \(3 \hat{i}-4 \hat{j}-4 \hat{k}\), \(2 \hat{i}-\hat{j}+\hat{k}\) and \(\hat{i}-3 \hat{j}-5 \hat{k}\) respectively form the vertices of a right-angled triangle. (NCERT)
(ii) If the vertices of a triangle are A (2, – 1, 1), B (1, – 3, – 5) and C (3, – 4, – 4), then prove by vector method that it is a right- angled triangle.
(iii) If the position vectors of the vertices of a triangle ABC are \(\hat{i}+2 \hat{j}+3 \hat{k}\), \(2 \hat{i}+3 \hat{j}+\hat{k}\) and \(3 \hat{i}+\hat{j}+2 \hat{k}\), then prove that ∆ABC is an equilateral triangle.
Solution:
Let A, B, C are the points whose position vectors are \(\vec{a}, \vec{b}, \vec{c}\)
i.e., \(\vec{a}=3 \hat{i}-4 \hat{j}-4 \hat{k}\) ;
\(\vec{b}=2 \hat{i}-\hat{j}+\hat{k}\)
and \(\vec{c}=\hat{i}-3 \hat{j}-5 \hat{k}\)

Clearly BC² = AB² + AC²
[∵ 41 = 35 + 6]
Hence pythagoras theorem is verified.
Thus the points A, B, C form the vertices of right angled ∆.

(ii) Let A, B, C are the points whose position vectors are latex]\vec{a}, \vec{b}, \vec{c}[/latex].

So points A, B and C form a triangle.
Clearly
AB² = CA² + CB²
[∵ 41 = 35 + 6]
Hence pythagoras theorem is verified. Thus the points A, B C form the vertices of right angled ∆.

(iii) Let O be the origin of reference.

Question 36.
The position vectors \(\vec{a}, \vec{b}, \vec{c}\) of three given points satisfy the relation \(4 \vec{a}-9 \vec{b}+5 \vec{c}=\overrightarrow{0}\). Prove that the three points are collinear.
Solution:
Let \(\vec{a}, \vec{b}, \vec{c}\) are position vectors of given points A, B and C respectively.
Since \(4 \vec{a}-9 \vec{b}+5 \vec{c}=\overrightarrow{0}\)
⇒ \(9 \vec{b}=4 \vec{a}+5 \vec{c}\)
⇒ \(\vec{b}=\frac{4 \vec{a}+5 \vec{c}}{4+5}\)

Thus point B divides CA internally in the ratio 4 : 5.
∴ point B lies on CA.
Thus A, B, C lies on same line and hence the given points A, B and C are collinear.

Question 38 (old).
If \(\vec{a}, \vec{b}\) are position vectors of A, B with reference to origin O, find the position vector of a point C in AB produced such that 2AC = 3AB.
Solution:
Given \(\vec{a}, \vec{b}\) are the position vectors of A
and B with reference to origin O.
∴ \(\overrightarrow{\mathrm{OA}}=\vec{a}\) ;
\(\overrightarrow{\mathrm{OB}}=\vec{b}\)
Given, 2AC = 3AB
⇒ \(2 \overrightarrow{\mathrm{AC}}=3 \overrightarrow{\mathrm{AB}}\)
⇒ \(2(\vec{c}-\vec{a})=3(\vec{b}-\vec{a})\)
where \(\vec{c}\) be the position vectors of point C.
⇒ \(2 \vec{c}=3 \vec{b}-3 \vec{a}+2 \vec{a}=3 \vec{b}-\vec{a}\)
⇒ \(\vec{c}=\frac{3 \vec{b}-\vec{a}}{2}\)

Interactive ML Aggarwal Class 12 ISC Solutions Chapter 9 Differential Equations MCQs engage students in active learning and exploration.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations MCQs

Choose the correct answer from the given four options in questions (1 to 53):

Question 1.
The degree of the differential equation \(\frac{d^2 y}{d x^2}+3\left(\frac{d y}{d x}\right)^2=x^2 \log \left(\frac{d^2 y}{d x^2}\right)\)
(a) 1
(b) 2
(c) 3
(d) Not defined
Solution:
(d) Not defined

Given diff. eqn. be, \(\frac{d^2 y}{d x^2}+3\left(\frac{d y}{d x}\right)^2=x^2 \log \left(\frac{d^2 y}{d x^2}\right)\)
Here R.H.S. of given diff. eqn. cannot be expressed as polynomial in derivative so degree of given diff. eqn. be not defined.

Question 2.
The degree of the differential equation \(\left(1+\left(\frac{d y}{d x}\right)^2\right)^{\frac{3}{2}}=\frac{d^2 y}{d x^2}\) is
(a) 4
(b) \(\frac{3}{2}\)
(c) 2
(d) not defined
Solution:
(c) 2

Given diff. eqn. be, \(\frac{d^2 y}{d x^2}=\left[1+\left(\frac{d y}{d x}\right)^2\right]^{3 / 2}\)
On squaring both sides ; we have
\(\left(\frac{d^2 y}{d x^2}\right)^2=\left[1+\left(\frac{d y}{d x}\right)^2\right]^3\)
The highest order derivative present in given diff. eqn. \(\frac{d^2 y}{d x^2}\) and its power is 2.
Thus the order of given diff. eqn. be 2 and degree 2.
Clearly it is a non linear differential eqn., as it contains terms like \(\left(\frac{d^2 y}{d x^2}\right)^2\) and \(\left(\frac{d y}{d x}\right)^2\).

Question 3.
The order and the degree of the differential equation \(\left(\frac{d y}{d x}\right)^5+3 x y\left(\frac{d^3 y}{d x^3}\right)^2+y^2\left(\frac{d^2 y}{d x^2}\right)^3\) = 0 respectively, are
(a) 2, 3
(b) 3, 2
(c) 3, 5
(d) 1, 5
Solution:
(b) 3, 2

The highest ordered derivative existing in given differential eqn. be \(\frac{d^3 y}{d x^3}\) so its order be 3.
Further, given diff. eqn. can be expressed as polynomial in derivatives.
So the highest exponent of \(\frac{d^3 y}{d x^3}\) which is 2 gives the degree of given diff. eqn.

Question 4.
The degree of the differential equation \(\left\{3-\left(\frac{d y}{d x}\right)^2\right\}^{5 / 3}=x^3\left(\frac{d^2 y}{d x^2}\right)\), is
(a) 2, 1
(b) 2, 2
(c) 2, 3
(d) 2, 5
Solution:
(c) 2, 3

Given differential eqn. be \(\left\{3-\left(\frac{d y}{d x}\right)^2\right\}^{5 / 3}=x^3\left(\frac{d^2 y}{d x^2}\right)\)
cubing on both sides ; we have
\(\left[3-\left(\frac{d y}{d x}\right)^2\right]^5=x^9\left(\frac{d^2 y}{d x^2}\right)^3\) …………….(1)
Here highest ordered derivative present in given diff. eqn (1) be \(\frac{d^2 y}{d x^2}\).
Hence, its order is 2 and also its power be 3.
∴ degree of given eqn (1) be 3.

Question 5.
The order and the degree of the differential equation \(\left(1+3 \frac{d y}{d x}\right)^{\frac{2}{3}}=4 \frac{d^3 y}{d x^3}\), respectively, are
(a) 3, \(\frac{3}{2}\)
(b) 3, 1
(c) 3, 2
(d) 3, 3
Solution:
(d) 3, 3

Given diff. eqn. be
\(\left(1+3 \frac{d y}{d x}\right)^{\frac{2}{3}}=4 \frac{d^3 y}{d x^3}\)
On cubing both sides; we have
\(\left(1+3 \frac{d y}{d x}\right)^2=\left(4 \frac{d^3 y}{d x^3}\right)^3\)
The highest ordered derivative existing in given diff. eqn. be \(\frac{d^3 y}{d x^3}\)
so its order be 3.
Further given diff. eqn. can be expressible as polynomial in derivatives.
Thus degree of given diff. eqn. be the highest exponent of \(\frac{d^3 y}{d x^3}\) which is 3.

Question 6.
The order and degree of the differential equation \(\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^{1 / 4}+x^{1 / 5}\) = 0
(a) 2, 4
(b) 2, 2
(c) 2, 3
(d) 2, not defined
Solution:
(a) 2, 4

Given diff. eqn be
\(\frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^{1 / 4}+x^{1 / 5}\) = 0
⇒ \(\frac{d^2 y}{d x^2}+x^{1 / 5}=-\left(\frac{d y}{d x}\right)^{1 / 4}\)
⇒ \(\left[\frac{d^2 y}{d x^2}+x^{y / 5}\right]^4=+\frac{d y}{d x}\)
which is polynomial in derivatives.
Also the highest ordered derivative existing in given diff. eqn be and its power be 4.
Thus the order of given differential eqn. be 2 and its degree be 4.

Question 7.
The order of the differential equation representing the family of curves y = ax + a³
(a) 1, 1
(b) 1, 3
(c) 1, 2
(d) 2, 3
Solution:
(b) 1, 3

y = ax + a³
∴ From given eqn ; we have
y = x \(\frac{d y}{d x}\) + (\(\frac{d y}{d x}\))³
which is diff. eqn. of order 1 and degree 3.

Question 8.
The order of the differential equation of the family of circles of radius r.
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(b) 2

Eqn. of families of circle with centre (h, k)
and radius r is given by
(x – a)² + (y – b)² = r² ……………..(1)
Diff. (1) w.r.t. x,. we have
2 (x – a) + 2 (y – b) y₁ = 0
⇒ (x – a) + (y – b) y₁ = 0 ……………….(2)
Diff. eqn (2) w.r.t. x; we have
1 + (y + b)y₂ + y₁² = 0
⇒ (y – b) = – \(\frac{\left(1+y_1^2\right)}{y_2}\)
∴ From (2); we have
x – a = – (y – b) y₁
= \(\frac{\left(1+y_1^2\right) y_1}{y_2}\)
∴ From eqn (1) ; we have
\(\frac{\left(1+y_1^2\right)^2 y_1^2}{y_2^2}+\frac{\left(1+y_1^2\right)^2}{y_2^2}\) = r²
clearly required order of duff. eqn. be 2.

Question 9.
Which of the following is a second order differential equation?
(a) (y’)² + x = y²
(b) y’ = y²
(c) y’y” + y = sin x
(d) y”’ + (y”)² + y = o.
Solution:
(c) y’y” + y = sin x

Clearly the highest ordered derivative existing in given duff. eqn. y’y” + y = sin x be y”. so its order be 2.

Question 10.
The differential equation of the family
of curves x² + y² – 2ay = 0, where a is arbitrary constant, is
(a) (x² – y²) \(\frac{d y}{d x}\) = 2xy
(b) 2 (x² + y²) \(\frac{d y}{d x}\) = xy
(c) 2 (x² – y²) \(\frac{d y}{d x}\) = xy
(d) (x² + y²) \(\frac{d y}{d x}\) = 2xy
Solution:
(a) (x² – y²) \(\frac{d y}{d x}\) = 2xy

Given diff. eqn. be, x² + y² – 2ay = 0 ……………(1)
Diff. eqn. (1) both sides w.r.t. x; we have
2x + 2y – 2a \(\frac{d y}{d x}\) = 0
⇒ (x + y \(\frac{d y}{d x}\)) = a \(\frac{d y}{d x}\)
∴ from (1) ; we have
x² + y² – 2y \(\frac{\left(x+y \frac{d y}{d x}\right)}{\frac{d y}{d x}}\) = 0
⇒ (x² + y²) \(\frac{d y}{d x}\) – 2xy – 2y² \(\frac{d y}{d x}\) = 0
⇒ (x² – y²) \(\frac{d y}{d x}\) = 2xy
which is the required differential equation.

Question 11.
The differential equation of the family of curves y² = 4a (x + a), where a is arbitrary constant, is
(a) y² = 4 \(\frac{d y}{d x}\) (x + \(\frac{d y}{d x}\))
(b) y \(\frac{d y}{d x}\) = 2a
(c) y \(\frac{d^2 y}{d x^2}\) + (\(\frac{d y}{d x}\))² = 0
(d) 2x \(\frac{d y}{d x}\) + y (\(\frac{d y}{d x}\))² = y
Solution:
(d) 2x \(\frac{d y}{d x}\) + y (\(\frac{d y}{d x}\))² = y

Given eq. be curve be
y² = 4a (x + a)
where a be an arbitrary constant.
Diff. eqn. (1) w.r.t. x ; we have
2y \(\frac{d y}{d x}\) = 4a
⇒ a = \(\frac{y}{2}\) \(\frac{d y}{d x}\)
∴ from (1) ; we have
y² = 2y \(\frac{d y}{d x}\) [x + \(\frac{y}{2} \frac{d y}{d x}\)]
⇒ y² = 2xy \(\frac{d y}{d x}\) + y² (\(\frac{d y}{d x}\))²

Question 12.
The differential equation for which y = a cos x + b sin x is a solution, is [NCERT Exemplar]
(a) \(\frac{d^2 y}{d x^2}\) + y = 0
(b) \(\frac{d^2 y}{d x^2}\) – y = 0
(c) \(\frac{d^2 y}{d x^2}\) + (a + b) y = 0
(d) \(\frac{d^2 y}{d x^2}\) + (a – b) y = 0
Solution:
(a) \(\frac{d^2 y}{d x^2}\) + y = 0

Given eqn. of solution be,
y = a cos x + b sin x ………………..(1)
Diff. eqn. (1) both sides w.r.t. x; we have
\(\frac{d y}{d x}\) = – a sin x + b cos x
Again diff. both sides w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = – a cos x – b sin x = – y
[using eqn. (1)]
⇒ \(\frac{d^2 y}{d x^2}\) + y = 0
which is the required differential equation.

Question 13.
y = ae^mx + be^-mx, where a, b are arbitrary constants satisfies which of the following differential equation?
(a) \(\frac{d y}{d x}\) + my = 0
(b) \(\frac{d y}{d x}\) – my = 0
(c) \(\frac{d^2 y}{d x^2}\) – m²y = 0
(d) \(\frac{d^2 y}{d x^2}\) + m²y = 0
Solution:
(c) \(\frac{d^2 y}{d x^2}\) – m²y = 0

Given y = ae^mx + be^-mx ………………(1)
where a and b are arbitrary constants
diff. eqn. (1) w.r.t. x ; we have
\(\frac{d y}{d x}\) = am e^mx – bm e^{– mx}
∴ \(\frac{d^2 y}{d x^2}\) = am² e^mx + bm² e^{– mx}
= m² [ae^mx + be^{– mx}]
= m² y [using (1)]

Question 14.
The differential equation obtained on eliminating A and B from y A cos ωt + B sin ωt, is
B sin mt, is
(a) y” + y’ = 0
(b) y” – ω²y = 0
(c) y” = – ω²y
(d) y” + y = 0
Solution:
(c) y” = – ω²y

Given y = A cos ωt + B sin ωt …………….(1)
y’ = – Aω sin ωt + Bω cos ωt
∴ y”= – Aω² cos ωt – Bω² sin ωt
= – ω² [A cos ωt + B sin ωt]
= – ω²y

Question 15.
The differential equation for the family of curves sin^-1 x + sin^-1 y = C, where C Is arbitrary constant, is
(a) \(\sqrt{1-x^2} d x+\sqrt{1-y^2} d y\) = 0
(b) \(\sqrt{1-x^2} d x-\sqrt{1-y^2} d y\) = 0
(c) \(\sqrt{1-x^2} d y+\sqrt{1-y^2} d x\) = 0
(d) \(\sqrt{1-x^2} d y-\sqrt{1-y^2} d x\) = 0
Solution:
(c) \(\sqrt{1-x^2} d y+\sqrt{1-y^2} d x\) = 0

Given sin^-1 x + sin^-1 y = C
where C be an arbitrary constant
Diff. eqn. (1) w.r.t. x;
\(\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-y^2}} \frac{d y}{d x}\) = 0
⇒ \(\frac{d x}{\sqrt{1-x^2}}+\frac{d y}{\sqrt{1-y^2}}\) = 0
⇒ \(\sqrt{1-y^2} d x+\sqrt{1-x^2} d y\) = 0

Question 16.
tan^-1 x + tan^-1 y = C, where C Is a parameter, is the general solution of the differential equation:
(a) \(\frac{d y}{d x}=\frac{1+y^2}{1+x^2}\)
(b) \(\frac{d y}{d x}=\frac{1+x^2}{1+y^2}\)
(c) (1+ x²) dy + (1 + y²) dx = 0
(d) (1 + x²) dy + (1 + y²) dy = 0
Solution:
(c) (1+ x²) dy + (1 + y²) dx = 0

Given tan^-1 x + tan^-1 y = C ……………………(1)
diff. eqn. (1) w.r.t. x ; we have
\(\frac{1}{1+x^2}+\frac{1}{1+y^2} \frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=-\frac{1+y^2}{1+x^2}\)

Question 17.
The differential equation representing the family of curves y² = 2c (x + √c) where c (> 0) is a parameter, is of
(a) order 2, degree 2
(b) order 1, degree 3
(c) order 1, degree 1
(d) order 1. degree 2
Solution:
(b) order 1, degree 3

Given y² = 2c (x + √c)
where c be an arbitrary constant
diff. eqn. (1) w.r.t. x ; we have
2yy’ = 2c
∴ from (1) ;
y² = 2yy’ [x + \(\)]
⇒ y = 2xy’ + 2y’ \(\sqrt{y y^{\prime}}\)
⇒ (y – 2xy’)² = 4y’² (yy’) = 4yy’³
Here the highest ordered derivative existing in given diff. eqn. be y’ so its order be 1.
Clearly the given diff. eqn. can be expressed as polynomial in derivatives.
Thus, the degree of given diff. eqn. be the highest exponent of y’ which is 3.

Question 18.
Which of the following ¡s not a homogeneous function of x and y?
(a) x² + 2y
(b) 2x – y
(c) cos² \(\left(\frac{y}{x}\right)+\frac{y}{x}\)
(d) sin x – cos y
Solution:
(d) sin x – cos y

x² + 2xy = x² [1 + 2(\(\frac{y}{x}\))]
= x² Φ (\(\frac{y}{x}\))
2 – xy = x [2 – \(\frac{y}{x}\)]
= x g (\(\frac{y}{x}\)) ;
cos² (\(\frac{y}{x}\)) + \(\frac{y}{x}\) = h (\(\frac{y}{x}\))
All of these functions are homogeneous in x and y.
Further sin x – sin y cant be expresed in the form of f (\(\frac{y}{x}\)).

Question 19.
Which of the following s a homogeneous differential equation?
(a) (2x – 3y + 5) dy – (3x + 2y – 4) dx = 0
(b) xy dx – (x³ – y³) dy = 0
(c) y² dx – (x² – xy + y²) dy = 0
(d) (y³ + 2xy) dx + 3x² dy = 0
Solution:
(c) y² dx – (x² – xy + y²) dy = 0

Clearly in option (c);
The given diff. eqn. can be written as;
\(\frac{d y}{d x}=\frac{y^2}{x^2-x y+y^2}\)
= \(\frac{\left(\frac{y}{x}\right)^2}{1-\frac{y}{x}+\left(\frac{y}{x}\right)^2}\)
= \(\phi\left(\frac{y}{x}\right)\)
which is a homogeneous function of degroe 0.

Question 20.
The number of solutions of \(\frac{d y}{d x}=\frac{y+1}{x-1}\), when y(1) = 2 is
(a) none
(b) one
(c) two
(d) infinite
Solution:
(b) one

Given \(\frac{d y}{d x}=\frac{y+1}{x-1}\) ……………..(1)
after variable separation, we have
\(\frac{d y}{y+1}=\frac{d x}{x-1}\) ;
on integrating

Question 21.
Integrating factor of the differential equation x \(\frac{d y}{d x}\) + y = 3x² is
(a) x
(b) log x
(c) \(\frac{1}{x}\)
(d) none of these
Solution:
(a) x

Given diff. eqn. can be written as;
\(\frac{d y}{d x}+\frac{y}{x}\) = 3x
which is L.D.E in y and is of the form
\(\frac{d y}{d x}\) + Py = Q ;
where P = \(\frac{1}{x}\) and Q = 3x
∴ I.F = e^{∫ P dx}
= \(e^{\int \frac{1}{x} d x}\)
= e^{log x}
= x

Question 22.
Integrating factor of x \(\frac{d y}{d x}\) – y = x⁴ -3x is
(a) x
(b) log x
(c) \(\frac{1}{x}\)
(d) – x
Solution:
Given diff. eqn. can be written as ;
\(\frac{d y}{d x}\) – \(\frac{y}{x}\) = 4x³ -3
which is of the form \(\frac{d y}{d x}\) + Py = Q
where P = – \(\frac{1}{x}\) and Q = x³ – 3
∴ I.F = e^{∫ P dx}
= \(\int_e-\frac{1}{x} d x\)
= e^{– log x}
= e ^{log x-1}
= x^-1
= \(\frac{1}{x}\)

Question 23.
Integrating factor of the differential equation cos x \(\frac{d y}{d x}\) + y sin x = 1, is
(a) sin x
(b) sec x
(c) tan x
(d) cos x
Solution:
(b) sec x

Given differential equation be,
cos x \(\frac{d y}{d x}\) + y sin x = 1
⇒ \(\frac{d y}{d x}+y \frac{\sin x}{\cos x}\) = sec x
which is L.D.E. of the form
\(\frac{d y}{d x}\) + Py = Q
where P = tan x
and Q = sec x
∴ I.F. = e^{∫ P dx}
= e^{∫ tan x dx}
= e^{– log cos x}
= \(e^{\log \frac{1}{\cos x}}\)
= \(\frac{1}{\cos x}\)
= sec x

Question 24.
Integratlng factor of the differential equation (1 – x²) \(\frac{d y}{d x}\) – xy = 1 is
(a) – x
(b) \(\frac{x}{1+x^2}\)
(c) \(\sqrt{1-x^2}\)
(d) \(\frac{1}{2}\) log(1 – x²)
Solution:
(c) \(\sqrt{1-x^2}\)

Given diff. eqn. can be written as;
\(\frac{d y}{d x}-\frac{x}{1-x^2} y=\frac{1}{\left(1-x^2\right)}\)
which is L.D.E in y and is of the form
\(\frac{d y}{d x}\) + Py = Q

Question 25.
The integrating factor of the differential equation (x log x) \(\frac{d y}{d x}\) + y = 2 log x, given by
(a) log (log x)
(b) e^x
(c) log x
(d) x
Solution:
(c) log x

Given diff. eqn. be,
(x log x) \(\frac{d y}{d x}\) + y = 2 log x
which is linear differential eqn.
and is of the form \(\frac{d y}{d x}\) + Py = Q
where P = \(\frac{1}{x \log x}\)
and Q = \(\frac{2}{x}\)
Now I.F = e^{∫ P dx}
= \(e^{\int \frac{1 / x}{\log x} d x}\)
= e^{log (log x)}
= log x

Question 26.
Integrating factor of the differential equation (1 – y²) \(\frac{d x}{d y}\) + xy = ay is
(a) \(\frac{1}{1-y^2}\)
(b) \(\frac{1}{\sqrt{y^2-1}}\)
(c) \(\frac{1}{y^2-1}\)
(d) \(\frac{1}{\sqrt{1-y^2}}\)
Solution:
(d) \(\frac{1}{\sqrt{1-y^2}}\)

Given diff. eqn. can be written as
\(\frac{d x}{d y}+\left(\frac{y}{1-y^2}\right) x=\frac{d y}{1-y^2}\)
which is L.D.E in x and is of the form
\(\frac{d x}{d y}\) + Py = Q

Question 27.
Integrating factor of the differential equation \(\frac{d y}{d x}\) + y = \(\frac{x^3+y}{x}\) is
(a) xe^x
(b) \(\frac{x}{e^x}\)
(c) \(\frac{e^x}{x}\)
(d) e^x
Solution:
(c) \(\frac{e^x}{x}\)

Given diff eqn. can be written as ;
\(\frac{d y}{d x}+\left(1-\frac{1}{x}\right) y\) = x²
which is L.D.E. in y and is of the form
\(\frac{d y}{d x}\) + Py = Q ;
where P = 1 – \(\frac{1}{x}\)
and Q = x²
∴ I.F. = e^{∫ P dx}
= \(e^{\int\left(1-\frac{1}{x}\right) d x}\)
= e^{x – log x}
= e^x e^{log x-1}
= e^x \(\frac{1}{x}\)

Question 28.
Solution of the differential \(\frac{d x}{x}\) + \(\frac{d y}{y}\) = 0 is
(a) \(\frac{1}{x}+\frac{1}{y}\) = C
(b) xy = C
(c) log x log y = C
(d) x + y = C
Solution:
(b) xy = C

Given diff eqn. can be written as ;
\(\frac{d x}{x}\) + \(\frac{d y}{y}\) = 0
On integrating both sides ; we have
\(\int \frac{d x}{x}+\int \frac{d y}{y}\) = log C
⇒ log |x| + log |y| = log C
⇒ log |xy| = log C
⇒ xy = ± C = A

Question 29.
The solution of the differential equation \(\frac{x d y}{d x}\) + 2y = x² is
(a) y = \(\frac{x^2}{4}\) + C
(b) y = \(\frac{x^2+C}{4 x^2}\)
(c) y = \(\frac{x^2+C}{x^2}\)
(d) y = \(\frac{x^4+C}{4 x^2}\)
Solution:
(d) y = \(\frac{x^4+C}{4 x^2}\)

Given differential equation can be written as;
\(\frac{d y}{d x}+\frac{2}{x} y\) + 2y = x²
which is first order L.D.E. in y and is of the form
\(\frac{x d y}{d x}\) + Py = Q ;
where P = \(\frac{2}{x}\) and Q = x
∴ I.F. = e^{∫ P dx}
= \(e^{\int \frac{2}{x} d x}\)
= e^{2 log |x|}
= e^{log |x|2}
= x²
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ y . x² = ∫ x . x² dx + C
⇒ x²y = \(\frac{x^4}{4}\) + C
⇒ y = \(\frac{x^4+C}{4 x^2}\) be the required solution.

Question 30.
Solution of the differential equation tan y sec² x dx + tan x sec² y dy is
(a) tan x+ tany = C
(b) tan x – tan y = C
(c) tan x tany = C
(d)tan x = C tan y
Solution:
(c) tan x tany = C

Given diff. eqn. can be written as ;
\(\frac{\sec ^2 x d x}{\tan x}+\frac{\sec ^2 y d y}{\tan y}\) = 0
On integrating ; we have
\(\int \frac{\sec ^2 x d x}{\tan x}+\int \frac{\sec ^2 y d y}{\tan y}\) = log C
⇒ log |tan x| + log |tan y| = log C
[∵ ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)| + C]
⇒ log |tan x tan y| = log C
⇒ tan x tan y = ± C = A

Question 31.
The solution of the differential equation \(\frac{d y}{d x}=\frac{1+y^2}{1+x^2}\) is
(a) y = tan^-1 x + C
(b) x = tan^-1 y + C
(c) tan (xy) = C
(d) y – x = C (1 + xy)
Solution:
(d) y – x = C (1 + xy)

Given \(\frac{d y}{d x}=\frac{1+y^2}{1+x^2}\)
After variable separation, we have
\(\frac{d y}{1+y^2}=\frac{d x}{1+x^2}\) ;
on integrating
\(\int \frac{d y}{1+y^2}-\int \frac{d x}{1+x^2}\) = A
⇒ tan^-1 y – tan^-1 x = A
⇒ tan^-1 \(\left(\frac{y-x}{1+x y}\right)\) = A
⇒ y – x = C (1 + xy) ; where tan A = C

Question 32.
The solution of th differential equation \(\frac{d y}{d x}=\sqrt{\frac{1-y^2}{1-x^2}}\) is
(a) sin^-1 y – sin^-1 x = C
(b) sin^-1 y + sin^-1 x = C
(c) sin^-1 (xy) = C
(d) none of these
Solution:
(a) sin^-1 y – sin^-1 x = C

After variable separation, we have
\(\frac{d y}{\sqrt{1-y^2}}=\frac{d x}{\sqrt{1-x^2}}\) ;
On integrating
\(\int \frac{d y}{\sqrt{1-y^2}}=\int \frac{d x}{\sqrt{1-x^2}}+\mathrm{C}\)
⇒ sin^-1 y – sin^-1 x = C

Question 33.
The general solution of e^x cos y dx – e^x sin y dy = 0 is
(a) e^x sin y = C
(b) e^x cos y = C
(c) e^x = C cos y
(d) e^x = C sin y
Solution:
(b) e^x cos y = C

Given diff. eqn. can be written as ;
e^x [dx – \(\frac{\sin y}{\cos y}\) dy] = 0
⇒ dx – \(\frac{\sin y}{\cos y}\) dy = 0
On integrating both sides ; we have
∫ dx = ∫ \(\frac{\sin y}{\cos y}\) dy + C
⇒ x = – log |cos y| + C
⇒ log |cos y| = C – x
⇒ |cos y| = e^{C – x}
= Ae^-x
⇒ e^x cos y = ± A = B

Question 34.
The solution of the equation (2y – 1) dx – (2x + 3) dy = 0 is
(a) \(\frac{2 x-1}{2 y+3}\) = C
(b) \(\frac{2 y+1}{2 x-3}\) = C
(c) \(\frac{2 x+3}{2 y-1}\) = C
(d) \(\frac{2 x-1}{2 y-3}\) = C
Solution:
(c) \(\frac{2 x+3}{2 y-1}\) = C

After variable separation, we have
\(\frac{d x}{2 x+3}-\frac{d y}{2 y-1}\) = 0 ;
On integrating

Question 35.
The general solution of \(\frac{d y}{d x}\) + y tan x = sec x is
(a) y sec x tan x + C
(b) y tan x = sec x + C
(c) x sec x = tan y + C
(d) none of these
Solution:
(a) y sec x tan x + C

Given diff. eqn. be
\(\frac{d y}{d x}\) + y tan x = sec x
which is linear diff. eqn. in y and is of the form
\(\frac{d y}{d x}\) + Py = Q ;
where P = tan x ; Q = sec x
∴ I.F. = e^{∫ P dx}
= e^{∫ tan x dx}
= \(e^{\int \frac{\sin x d x}{\cos x}}\)
= e^{– log |cos x|}
= e^{log |cos x|-1}
= \(\frac{1}{cos x}\)
= sec x
and general solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ y . sec x = ∫ sec² x dx + C
= tan x + C

Question 36.
The solution of the differential equation \(\frac{d y}{d x}+\frac{2 x y}{1+x^2}=\frac{1}{\left(1+x^2\right)^2}\) is
(a) y log (1 + x²) = tan^-1 x + C
(b) y (1 + x²) = tan^-1 x + C
(c) y (1 + x²) sin^-1 x + C
(d) \(\frac{y}{1+x^2}\) = tan^-1 x + C
Solution:
(b) y (1 + x²) = tan^-1 x + C

Given diff. eqn. be,
\(\frac{d y}{d x}+\left(\frac{2 x}{1+x^2}\right) y=\frac{1}{\left(1+x^2\right)^2}\)
which is L.D.E my and is of the form
\(\frac{d y}{d x}\) + Py = Q;
where P = \(\frac{2 x}{1+x^2}\)
and Q = \(\frac{1}{\left(1+x^2\right)^2}\)
Here I.F = e^{∫ P dx}
= \(e^{\int \frac{2 x}{1+x^2} d x}\)
= e^{log (1 + x2)}
= 1 + x²
and general solution be given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ y . (1 + x²) = ∫ \(\frac{1}{\left(1+x^2\right)^2}\) (1 + x²) dx + C
= tan^-1 x + C

Question 37.
The general solution of the differential equation \(\frac{d y}{d x}\) + y cot x = cosec x, is
(a) x + y sin x = C
(b) x + y cos x = C
(c) y + x (sin x + cos x) = C
(d) y sin x = x + C
Solution:
(d) y sin x = x + C

Given \(\frac{d y}{d x}\) + y cot x = cosec x
which is L.D.E. of the form \(\frac{d y}{d x}\) + Py = Q
where P = cot x & Q = cosec x
∴ I.F. = e^{∫ P dx}
= e^{∫ cot x dx}
= e^{log sin x}
= sin x
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + c
y . sin x = ∫ cosec x . sin x dx + c
= x + c

Question 38.
The solution of the differential equation \(\frac{d y}{d}-\frac{y(x+1)}{x}\) = 0 is given by
(a) y = xe^{x + c}
(b) x= ye^x
(c) y = x + C
(d) xy = e^x + C
Solution:
(a) y = xe^{x + c}

Given diff. eqn be \(\frac{d y}{d}- y \frac{(x+1)}{x}\) = 0
⇒ \(\frac{1}{y} d y-\left(\frac{x+1}{x}\right) d x\) = 0 ;
on integrating
⇒ log y – (x + log x) = C
⇒ log \(\frac{y}{x}\) = x + C
⇒ \(\frac{y}{x}\) = x + C
⇒ \(\frac{y}{x}\) = e^{x + C}
⇒ y = xe^{x + C} be the required solution.

Question 39.
The general solution of the differential equation \(\frac{d y}{d x}\) = e^{x + y}, is
(a) e^x + e^-y = C
(b) e^x + e^y = C
(c) e^-x + e^y = C
(d) e^-x + e^-y = C
Solution:
(a) e^x + e^-y = C

Given diff. eqn be,
\(\frac{d y}{d x}\) = e^{x + y}
⇒ \(\frac{d y}{d x}\) = e^-x . e^y
⇒ e^-y dy = e^x dx
on integrating; we have
e^-y = e^x + c
⇒ e^-y + e^-x = A be the required solution.

Question 40.
General solution of the differential equation(1 + x) y dx + (1 – y) x dy = 0 is
(a) log xy + x – y = C
(b) log xy – x + y = C
(c) log (\(\frac{x}{y}\)) + x + y = C
(d) log (\(\frac{x}{y}\)) – x + y = C
Solution:
(a) log xy + x – y = C

Given differential eqn. can be written s;
\(\frac{(1+x)}{x} d x+\frac{(1-y)}{y} d y\) = 0
On integrating both sides ; we have
\(\int\left(\frac{1}{x}+1\right) d x+\int\left(\frac{1}{y}-1\right) d y\) = C
⇒ log |x| + x + log |y| – y = C
⇒ log |xy| + x – y = C

Question 41.
Solution of the differential equation \(\frac{d y}{d x}\) = e^{x – y} + x² e^-y is
(a) y = e^{x – y} – x² e^-y + C
(b) e^x + e^y = \(\frac{x^3}{3}\) + C
(c) e^y – e^x = \(\frac{x^3}{3}\) + C
(d) e^x – e^y = \(\frac{x^3}{3}\) + C
Solution:
(c) e^y – e^x = \(\frac{x^3}{3}\) + C

Given \(\frac{d y}{d x}\) = e^{x – y} + x² e^-y
= e^-y (e^x + x²)
On integrating both sides;
⇒ ∫ e^y dy = ∫ (e^x + x²) dx + C
⇒ e^y = e^x + \(\frac{x^3}{3}\) + C

Question 42.
General solution of the differential equation \(\frac{d y}{d x}\) = \(e^{\frac{x^2}{2}}\) + xy is
(a) Y = (C – x) \(e^{\frac{x^2}{2}}\)
(b) y = (x + C) \(e^{\frac{x^2}{2}}\)
(c) y = C\(e^{-\frac{x^2}{2}}\)
(d) y = C
Solution:
(b) y = (x + C) \(e^{\frac{x^2}{2}}\)

Given diff. eqn. be,
\(\frac{d y}{d x}\) = \(e^{\frac{x^2}{2}}\) + xy
⇒ \(\frac{d y}{d x}\) – xy = \(e^{\frac{x^2}{2}}\)
which is L.D.E in y and is of the form
\(\frac{d y}{d x}\) + Py = Q ;
where P = – x ; Q = \(e^{\frac{x^2}{2}}\)
∴ I.F. = e^{∫ P dx}
= e^{∫ – x dx}
= \(e^{-\frac{x^2}{2}}\)
and general solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ y\(e^{-\frac{x^2}{2}}\) = ∫ \(\int e^{\frac{x^2}{2}-\frac{x^2}{2}}\) dx + C
= x + C
⇒ y = (x + C) \(e^{\frac{x^2}{2}}\)

Question 43.
General solution of the differential equation \(\frac{d y}{d x}\) = 2x e^{x2 – y} is
(a) e^{– y} + C
(b) e^y = e^x2 + C
(c) e^{x2 – y} = C
(d) e^{x2 + y} = C
Solution:
(b) e^y = e^x2 + C

Given \(\frac{d y}{d x}\) = 2x e^{x2 – y}
= 2x e^x2 e^-y
⇒ e^y dy = 2x e^x2 dx
[after variable separation]
On integrating both sides we have
∫ e^y dy = ∫ 2x e^x2 dx + C
⇒ e^y = ∫ e^t dt + C
[where x² = t
⇒ 2x dx = dt]
⇒ e^y = e^t + C
⇒ e^y = e^x2 + C

Question 44.
General solution of the differential equation (e^x + 1) y dy = (y + 1) e^x dx is
(a) y + 1 = C (e^x + 1)
(b) y + 1 = e^x + 1 + C
(c) y = log(C (y + 1) (e^x + 1))
(d) none of these
Solution:
(c) y = log(C (y + 1) (e^x + 1))

Given (e^x + 1) y dy = (y + 1) e^x dx
After variable separation, we have

[where e^x = t
= e^x dx = dt]
⇒ y – log |y + 1| = log |t + 1| + cos C
⇒ y – log |y + 1| = log |e^x + 1| + log c
⇒ y = log {|y + ||e^x + 1| C}
⇒ y = log (C (y + 1) (e^x + 1))

Question 45.
General solution of the differential equation x \(\frac{d y}{d x}\) + y = e^x is
(a) y = \(\frac{e^x}{x}+\frac{C}{x}\)
(b) y = x e^x + Cx
(c) y = xe^x + C
(d) x = \(\frac{e^y}{y}+\frac{c}{y}\)
Solution:
Given diff. eqn. can be written as
\(\frac{d y}{d x}+\frac{y}{x}=\frac{e^x}{x}\)
which is L.D.E in y and is of the form
\(\frac{d y}{d x}\) + Py = Q ;
where P = \(\frac{1}{x}\)
and Q = \(\frac{e^x}{x}\)
I.F. = e^{∫ P dx}
= \(e^{\int \frac{1}{x} d x}\)
= e^{log x} = x
and general solution be given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
y . x = ∫ \(\frac{e^x}{x}\) . x dx + C
= e^x + C
⇒ y = \(\frac{e^x}{x}+\frac{c}{x}\)

Question 46.
Solution of the differential equation \(\frac{d y}{d x}\) + ay = e^mx is
(a) (a + m) y = e^mx + Ce^{– ax}
(b) y = e^mx + Ce^-ax
(c) (a + m) y = e^mx + C
(d) ye^ax = me^mx + C
Solution:
(a) (a + m) y = e^mx + Ce^{– ax}

Given diff. eqn. be,
\(\frac{d y}{d x}\) + ay = e^mx
which is L.D.E in y and is of the form
\(\frac{d y}{d x}\) + Py = Q ;
where P = a ; Q = e^mx
∴ I.F. = e^{∫ P dx}
= e^{∫ a dx}
= e^ax
and soln, be given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} + C
y . e^ax = ∫ e^mx . e^ax dx + C
= ∫ e^{(m + a) x} dx + C
= \(\frac{e^{(m+a) x}}{m+a}\) + C
⇒ (m + a) y = e^mx + C’ e ^ax

Question 47.
Solution of the differential equation \(\frac{d y}{d x}+\frac{y}{x}\) = sin x is
(a) x (y + cos x) = sin x + C
(b) x(y – cos x) = sin x + C
(c)x (y + cos x) = cos x + C
(d) none of these
Solution:
(a) x (y + cos x) = sin x + C

Given diff. eqn be,
\(\frac{d y}{d x}+\frac{y}{x}\) = sin x
which is L.D.E. of the from \(\frac{d y}{d x}\) + Py = Q
where P = \(\frac{1}{x}\) and Q = sin x
∴ I.F. = e^{∫ P dx}
= \(e^{\int \frac{1}{x} d x}\)
= e^{log x}
= x
and solution is given by
y . e^{∫ P dx} = Q . ∫ e^{∫ P dx} dx + c
⇒ y.x = ∫ sin x . x dx + c
⇒ xy = [- x cos x + sin x] + c
⇒ x (y + cos x) = sin x + c
which is the reqd. solution.

Question 48.
General solution of the differential equation cos x sin y dx + sin x cosy dy = 0 is
(a) sin x = C sin y
(b) sin x siny = C
(e) sin x + sin.y C
(d) cos x cosy = C
Solution:
(b) sin x siny = C

Given diff. eqn. can be written as;
\(\frac{\cos x d x}{\sin x}+\frac{\cos y d y}{\sin y}\) = 0
On integrating both sides ; we have
\(\int \frac{\cos x}{\sin x} d x+\int \frac{\cos y}{\sin y} d y\) = log C
⇒ log |sin x| + log |sin y| = log C
⇒ log |sin x sin y| = log C
⇒ sin x sin y = ± C = A

Question 49.
The solution of the differential equation \(\frac{d y}{d x}\) = y = e^-x, y(0) = 0, is
(a) y = e^x (x – 1)
(b) y = xe^-x
(c) y = xe^-x + 1
(d) y = (x + 1) e^-x
Solution:
(b) y = xe^-x

Given diff. eqn. be,
\(\frac{d y}{d x}\) + y = e^-x
which is L.Ð.E in y and is of the form
\(\frac{d y}{d x}\) + Py = Q
where P = 1 and Q = e^-x
∴ I.F = e^{∫ P dx}
= e^{∫ 1 dx}
= e^x
and soln. be given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ y e^x = ∫ e^{– x} .e^x dx + C
= ∫ 1 dx + C
= x + C ………………(1)
Given y(0) = 0 i.e.
when x = 0
⇒ y = 0
∴ from (1) ;
0 = (0 + C)
⇒ C = 0
Thus eqn. (1) becomes;
y = x e^{– x}

Question 50.
The solution of the differential equation \(\frac{d y}{d x}\) – y = 1, y(0) = 1, is
(a) xy = e^{– x}
(b) xy = – e^{– x}
(c) xy = – 1
(d) y = 2 e^x – 1
Solution:
(d) y = 2 e^x – 1

Given diff. eqn. be
\(\frac{d y}{d x}\) – y = 1
which is L.D.E in y and is of the form
\(\frac{d y}{d x}\) + Py = Q ;
where P = – 1 and Q = 1
∴ I.F = e^{∫ P dx}
= e^{– ∫ dx}
= e^{– x}
and general soln, be given by
⇒ y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ ye^{– x} = ∫ 1 . e^{– x}dx + C
⇒ ye^{– x} = \(\frac{e^{-x}}{-1}\) + C
⇒ y = – 1 + Ce^x …………………(1)
Given y (0) = 1
i.e. When x = 0 ⇒ y = 1
⇒ 1 = – 1 + C
⇒ C = 2
from (1) ;
y = – 1 + 2e^x

Question 51.
Solution of the differential equation x dy -y dx = 0 represents a family of
(a) rectangular hyperbolas
(b) parabolas with vertex at origin
(c) straight lines passing through origin
(d) circles whose centre is at origin
Solution:
(c) straight lines passing through origin

Given diff. eqn. can be written as:
\(\frac{d y}{y}-\frac{d x}{x}\) = 0 ;
On integrating ; we have
\(\int \frac{d y}{y}-\int \frac{d x}{x}\) = log C
⇒ log y – log x = log C
⇒ log \(\frac{y}{x}\) = log C
⇒ y = Cx
which represents a family of straight lines passing through origin.

Question 52.
Solution of the differential equation y \(\frac{d y}{d x}\) + x = 0 represents a family of
(a) hyperbola
(b) parabolas
(c) ellipses
(d) circles
Solution:
(d) circles

Given diff. eqn. can be written as;
y dy + x dy = 0 ;
on integrating ; we have
∫ y dy + ∫ x dx = 1
⇒ \(\frac{y^2}{2}+\frac{x^2}{2}=\frac{C}{2}\)
⇒ x² + y² = C
which represents a family of circles.

Question 53.
The solution of the differential equation 2x \(\frac{d y}{d x}\) – y = 3 represents
(a) circles
(b) straight lines
(c) ellipses
(d) parabolas
Solution:
(d) parabolas

Given diff. eqn be, 2x \(\frac{d y}{d x}\) – y = 3
⇒ \(\frac{d y}{d x}-\frac{y}{2 x}=\frac{3}{2 x}\)
which is L.D.E. of the form \(\frac{d y}{d x}\) + Py = Q

on squaring; we have
(y + 3)² = c²x,
which represents a right handed parabolas with vertex (0, – 3).

Utilizing ML Aggarwal Class 12 ISC Solutions Chapter 9 Differential Equations Chapter Test as a study aid can enhance exam preparation.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Chapter Test

Question 1.
Verify that the function y = e^ax (c₁ cos bx + c₂ sin bx), where c₁, c₂ are arbitrary constants is a solution of the differential equation \(\frac{d^2 y}{d x^2}\) – 2a \(\frac{d y}{d x}\) + (a² + b²) y = 0. (NCERT)
Solution:
Given,
y = e^ax (c₁ cos bx + c₂ sin bx) …………….(1)
Diff. (1) w.r.t. x we have
\(\frac{d y}{d x}\) = ae^ax (c₁ cos bx + c₂ sin bx) + e^ax (- c₁ b sin bx + c₂ b cos bx)
⇒ \(\frac{d y}{d x}\) = ay + be^ax (- c₁ sin bx + c₂ cos bx)
Diff. eqn. (2) w.r.t. x ; we have
\(\frac{d^2 y}{d x^2}\) = a \(\frac{d y}{d x}\) + b {e^ax {- c₁ b cos bx – c₂ b sin bx) + (- c₁ sin bx + c₂ cos bx) ae^ax}
⇒ \(\frac{d^2 y}{d x^2}\) = a \(\frac{d y}{d x}\) – b²y + a \(\frac{d y}{d x}\) – a²y
[using eqn. (1) and (2)]
⇒ \(\frac{d^2 y}{d x^2}\) – 2a \(\frac{d y}{d x}\) + (a² + b²) y = 0
which is the given diff. eqn.
Hence, y = e^ax (C₁ cos bx + C₂ sin bx) be the solution of given diff. eqn.

Question 1 (old).
Show that y = \(\frac{c-x}{1+c x}\), is arbitrary constant, is a solution of the differential equation (1 + x²) \(\frac{d y}{d x}\) + (1 + y²) = 0.
Solution:
Given y = \(\frac{C-x}{1+C x}\)
⇒ y (1 + Cx) = C – x
⇒ C (xy – 1) = – x – y
⇒ C = \(\frac{x+y}{1-x y}\) ……………..(1)
Diff. (1) w.r.t. x; we have
\(\frac{(1-x y)\left(1+\frac{d y}{d x}\right)+(x+y)\left(x \frac{d y}{d x}+y\right)}{(1-x y)^2}\) = 0
⇒ (1 + x²) \(\frac{d y}{d x}\) + 1 + y² = 0, which is given diff. eqn.
Thus, y = \(\frac{c-x}{1+c x}\) be the required solution of given diff. eqn.

Question 2.
Form the differential equation of the family of curves represented by \(\sqrt{1-x^2}+\sqrt{1-y^2}\) = a (x – y), a being parameter.
Solution:
Given family of curves represented by
\(\sqrt{1-x^2}+\sqrt{1-y^2}\) = a (x – y)
put x = sin θ
⇒ θ = sin^-1 x
and y = sin Φ
Φ = sin^-1 y
∴ From (1) ; we have
\(\sqrt{1-\sin ^2 \theta}+\sqrt{1-\sin ^2 \theta}\) = a (sin θ – sin Φ)
⇒ cos θ + cos Φ = a (sin Φ – sin Φ)
⇒ \(2 \cos \frac{\theta+\phi}{2} \cos \frac{\theta-\phi}{2}=2 a \cos \frac{\theta-\phi}{2} \sin \frac{\theta+\phi}{2}\)
⇒ cot \(\frac{\theta+\phi}{2}\) = a
⇒ θ + Φ = 2 cot^-1 a
⇒ sin^-1 x + sin^-1 y = 2 cot^-1 a
Diff. both sides w.r.t. x; we have
\(\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-y^2}} \frac{d y}{d x}\) = 0
which gives the required differential equation.

Question 3.
Obtain the differential equation of the family of straight lines which are at a fixed distance p from the origin.
Solution:
Now, equation of family of straight lines which are at a fixed distance p from the origin is given by
x cos α + y sin α = p ………………..(1)
where α be the parameter
On differentiating eqn. (1) w.r.t. x; we have
cos α + sin α \(\frac{d y}{d x}\) = 0 ……………..(2)
eqn. (1) – x × eqn. (2) ; we have
sin α (y – x \(\frac{d y}{d x}\)) = p ……………….(3)
eqn. (1) × \(\frac{d y}{d x}\) – eqn. (2) × y ; we have
(x \(\frac{d y}{d x}\) – y) cos α = p \(\frac{d y}{d x}\) ………………(4)
On squaring and adding eqn. (3) and eqn. (4) ; we have
(xy₁ – y)² (cos² α + sin² α) = p² (1 + (\(\frac{d y}{d x}\))²]
⇒ (y – xy₁)² = p² (1 + y₁²)
which gives the required differential eqn.

Question 4.
(tan² x + 2 tan x + 5) = 2 (1 + tan x) sec² x.
Solution:
Given, (tan² x + 2 tan x + 5) = 2 (1 + tan x) sec² x
⇒ dy = \(\frac{2(1+\tan x) \sec ^2 x}{\tan ^2 x+2 \tan x+5}\) dx
[after variable separation]
On integrating ; we have
∫ dy = ∫ \(\frac{2(1+\tan x) \sec ^2 x d x}{\tan ^2 x+2 \tan x+5}\) + C ……………….(1)
put tan² x + 2 tan x + 5 = t
⇒ (2 tan x sec² x + 2 sec² x) dx = dt
⇒ 2 sec² x (1 + tan x) dx = dt

Question 5.
(i) x (y² – a) dx + y (1 + x) dy = 0
(ii) cos x (1 + cos y) dy – sin y (1 + sin x) dx = 0
Solution:
(i) Given, x (y² – a) dx + y (1 + x) dy = 0
⇒ \(\frac{x d x}{1+x}+\frac{y d y}{y^2-a}\) = 0
[after variable separation]
On integrating ; we have
\(\int \frac{1+x-1}{1+x} d x+\int \frac{y d y}{y^2-a}\) = C
⇒ \(\int\left[1-\frac{1}{1+x}\right] d x+\frac{1}{2} \int \frac{2 y d y}{y^2-a}\) = C
⇒ x – log |1 + x| + \(\frac{1}{2}\) log |y² – a| = C
which gives the required solution.

(ii) Given, cos x (1 + cos y) dy – sin y (1 + sin x) dx = 0

Question 5 (old).
Find the differential equation of all parabolas having their axis of symmetry as the x-axis.
Solution:
The equation of all parabolas having axis of symmetry as the x-axis be given by
(y – 0)² = 4a (x – h) ……………..(1)
where a, h are arbitrary constants
Diff. eqn. (1) both sides w.r.t. x; we have
2yy₁ = 4a
⇒ yy₁ = 2a ……………..(2)
again differentiating eqn. (2) w.r.t. x, we get
yy₂ + y₁ = 0
which gives the required differential equation.

Question 6.
(i) \(\frac{d y}{d x}\) = x⁵ tan^-1 (x³)
(ii) a (x \(\frac{d y}{d x}\) + 2y) = xy \(\frac{d y}{d x}\)
Solution:
(i) Given, \(\frac{d y}{d x}\) = x⁵ tan^-1 (x³)
⇒ dy = x⁵ tan^-1 (x³) dx
On integrating ; we have
∫ dy = ∫ x⁵ tan^-1 (x³) dx
put x³ = t
⇒ 3x² dx = dt
y = ∫ t tan^-1 t \(\frac{d t}{3}\) + C

(ii) Given a (x \(\frac{d y}{d x}\) + 2y) = xy \(\frac{d y}{d x}\)
⇒ x (a – y) \(\frac{d y}{d x}\) = – 2 ay
\(\frac{(a-y)}{a y}=-2 \frac{d x}{x}\)
On integrating ; we have
\(\frac{1}{a} \int\left[\frac{a}{y}-1\right] d y=-2 \int \frac{d x}{x}+\mathrm{C}\)
⇒ \(\frac{1}{a}\) [a log |y| – y] = – 2 log |x| + C
⇒ log |x²y| = C + \(\frac{y}{a}\)
⇒ a log |yx²| = y + A
which gives the required general solution.

Question 7.
cos x cos y \(\frac{d y}{d x}\) + sin x sin y = 0.
Solution:
Given, cos x cos y \(\frac{d y}{d x}\) + sin x sin y = 0
\(\frac{\cos y}{\sin y} d y+\frac{\sin x}{\cos x} d x\) = 0
On integrating ; we have
\(\int \frac{\cos y}{\sin y} d y+\int \frac{\sin x}{\cos x} d x\) = log C
⇒ log |sin y| – log |cos x| = log C
⇒ |sin x| = C |cos x|
⇒ sin y = A cos x ; where ± C = A
which gives the required general solution.

Question 8.
Find the particular solution of (1 – x) dy – (1 + y) dy = 0, it being given that y = 4 when x = 2.
Solution:
Given, (1 – x) dy – (1 + y) dy = 0
⇒ \(\frac{d y}{1+y}-\frac{d x}{1-x}\) = 0
On integrating; we have
\(\int \frac{d y}{1+y}-\int \frac{d x}{1-x}\) = log C
⇒ log (1 + y) + log (1 – x) = log C
⇒ (1 + y) (1 – x) = C
Thus, eqn. (1) gives the general solution of given differential eqn.
Given y = 4 when x = 2
∴ from (1) ; we have
5(- 1) = C
⇒ C = – 5
Thus from eqn. (1) ; we have
(1 + y) (1 – x) + 5 = 0
which gives the required particular solution of given diff. eqn.

Question 9.
Find the particular solution of the differential equation \(e^{\frac{d y}{d x}}\) = x + 1, given that y = 3 when x = 0.
Solution:
Given \(e^{\frac{d y}{d x}}\) = log (x + 1)
\(\frac{d y}{d x}\) = log (x + 1) dx ;
on integrating ∫ dy = ∫ log (x + 1) . 1 dx
⇒ y = log (x + 1) . x – \(\int \frac{1}{x+1}\) x dx
⇒ y = x log |(x + 1)| – ∫ \(\left[1-\frac{1}{x+1}\right]\) dx + c
⇒ y = x log (x + 1) – x + log (x + 1) + C
⇒ y = (x + 1) log |(x + 1)| – x + c ………….(1)
Given y = 3 when x = 0
∴ from (1) ; we have
3 = log 1 – 0 + c
⇒ c = 3
From (1) ; we have
∴ y = (x + 1) log |x + 1| – x + 3
which is the required solution.

Question 10.
Find the equation of the curve passing through the point (0, \(\frac{\pi}{4}\)) whose differential equation is sin x cos y dx + cos x sin y dy = 0. (NCERT)
Solution:
Given differential equation be,
sin x cos y dx + cos x sin y dy = 0
⇒ \(\frac{\sin x}{\cos x}\) dx + \(\frac{\sin y}{\cos y}\) dy = 0
On integrating ; we have
\(\int \frac{\sin x d x}{\cos x}+\int \frac{\sin y d y}{\cos y}\) = – log C
⇒ – log |cos x| – log |cos y| = – log C
⇒ log |cos x cos y| = log C
⇒ cos x cos y = ± C = A ……………(1)
Since the curve (1) passes through (0, \(\frac{\pi}{4}\)).
i.e. When x = 0, y = \(\frac{\pi}{4}\)
∴ from (1) ;
\(1 \times \frac{1}{\sqrt{2}}\) = A
⇒ A = \(\frac{1}{\sqrt{2}}\)
Thus eqn. (1) gives ; cos x cosy = \(\frac{1}{\sqrt{2}}\) which gives the required eqn. of curve.

Question 11.
In a bank principal increases at the rate of 5 % per year. In how many years Rs. 1000 double itself. (NCERT)
Solution:
Let P be the principal at any instant t
Then \(\frac{d \mathrm{P}}{d t}=\frac{5}{100} \mathrm{P}\)
⇒ \(\frac{\mathrm{dP}}{\mathrm{P}}=\frac{d t}{20}\)
on integrating; we have
log P = \(\frac{1}{20}\) t + c …………….(1)
initially at t = 0 ; P = 1000
∴ from (1) ; we have
log 1000 = \(\frac{1 \times 0}{20}\) + c
⇒ c = log 1000
∴ from eqn (1) ; we have
log P = \(\frac{t}{20}\) + log (1000)
log \(\frac{\mathrm{P}}{1000}=\frac{t}{20}\) …………………..(2)
Let t = t₁ when P = 2000
∴ from eqn. (2); we have,
log \(\left(\frac{2000}{1000}\right)=\frac{t_1}{20}\)
⇒ t₁ = 20 log_e 2
Hence, the amount will be double after 20 log_e 2 years.

Question 12.
(i) x (x – y) dy + y² dx = 0
(ii) x³ \(\frac{d y}{d x}\) = y³ + y² \(\sqrt{y^2-x^2}\), x > 0
Solution:
(i) Given, x (x – y) dy + y² dx = 0

(ii) Given, x³ \(\frac{d y}{d x}\) = y³ + y² \(\sqrt{y^2-x^2}\)

Question 13.
(i) \(\left(\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right) \frac{d x}{d y}\) = 1, x > 0
(ii) e^-y sec² y dy = dx + x dy
Solution:
(i) Given, \(\left(\frac{e^{-2 \sqrt{x}}}{\sqrt{x}}-\frac{y}{\sqrt{x}}\right) \frac{d x}{d y}\) = 1

(ii) Given, e^-y sec² y dy = dx + x dy
⇒ (e^-y sec² y – x) dy = dx
⇒ \(\frac{d x}{d y}\) + y = e^-y sec² y
which is L.D.E. in x and is of the form
\(\frac{d x}{d y}\) + Px = Q
where P = 1 and Q = e^-y sec² y
∴ I.F. = e^{∫ P dy}
= e^{∫ dy}
= e^y
and solution is given by
x . e^{∫ P dy} = ∫ Q . e^{∫ P dy} dy + C
⇒ x . e^y = ∫ e^{– y} sec² y e^y dy + C
⇒ x . e^y = tan y + C
⇒ x = e^{– y} (tan y + C)

Question 14.
(i) (2x – 10y³) dy + y dx = 0, y ≠ 0
(ii) (x + tan y) dy = sin 2y dx
Solution:
(i) Given, (2x – 10y³) dy + y dx = 0
⇒ \(\frac{d x}{d y}\) + \(\frac{2 x}{y}\) – 10 y² = 0
⇒ \(\frac{d x}{d y}\) + \(\frac{2 x}{y}\) = 10 y²
which is linear in x and its of the form
⇒ \(\frac{d x}{d y}\) + Px = Q ;
where P = \(\frac{2}{y}\)
Q = 10 y²
∴ I.F. = e^{∫ P dy}
= \(e^{\int \frac{2}{y} d y}\)
= e^{2 log |y|}
= e^{log y2}
= y²
and solution is given by
x . e^{∫ P dy} = ∫ Q . e^{∫ P dy} dy + C
⇒ x . y² = ∫ 10 y² × y² dy + C
⇒ xy² = 10 + C
⇒ xy² = 2y⁵ + C, which gives the required solution.

(ii) Given (x + tan y) dy = sin 2y dx

Question 14 (old).
In a bank principal increases at the rate of r % per year. Find the value of r if Rs. 100 double itself in 10 years (given log_e 2 = 0.6931). (NCERT)
Solution:
Let P be the pnncipal at any instant t
Then \(\frac{d \mathrm{P}}{d t}=\frac{r}{100} \mathrm{P}\)
⇒ \(\frac{d \mathrm{P}}{P}=\frac{r}{100}\) dt ;
on integrating ; we get
⇒ log P = \(\frac{r}{100}\) t + c ……………..(1)
initially at t = 0, P = P₀
∴ from (1) ; we have
log P₀ = c
∴ from (1) ; we have
log P = \(\frac{r t}{100}\) + log P₀
⇒ \(\left(\frac{p}{P_0}\right)=\frac{r t}{100}\) …………….(2)
Given P₀ = 100 ; P = 200 at t = 10
∴ from (2) ; we have
⇒ \(\left(\frac{200}{100}\right)=\frac{r \times 10}{100}\)
⇒ log 2 = \(\frac{r}{10}\)
⇒ r = 10 × log 2 = 10 × 0.6931 = 6.931
Thus the required rate percent be 6.931% per annum.

Question 15.
Find the particular solutions of the following differential equations :
(i) \(\frac{d y}{d x}\) = 2xy, given that y = 1 when x = 0.
(ii) 2x² \(\frac{d y}{d x}\) – 2xy + y² = 0, given that y = e when x = e.
(iii) (xy – y²) dx – x² dy = 0, given that y = 1 when x = 1.
(iv) (x² + 1) \(\frac{d y}{d x}\) – 2xy = (x⁴ + 2x² + 1) cos x, given that y = 0 When x = 0.
(v) \(\sqrt{1-y^2}\) dx = (sin^-1 y – x) dy, given that y = 0 when x = 0.
(vi) \(x e^{\frac{y}{x}}-y \sin \left(\frac{y}{x}\right)+x \sin \left(\frac{y}{x}\right) \frac{d y}{d x}\) = 0, given that y = 0 when x = 1.
Solution:
(i) Given, \(\frac{d y}{d x}\) = 2xy
⇒ \(\frac{1}{y}\) dy = 2x dx
On integrating ; we have
log |y| = x² + C ………………..(1)
given, y = 1 when x = 0
∴ from (1) ;
0 = 0 + C
⇒ C = 0
Thus eqn. (1) gives ;
log |y| = x²
⇒ y = e^x2
[∵ y > 0]
which gives the required particular solution.

(ii) Given, 2x² \(\frac{d y}{d x}\) – 2xy + y² = 0

⇒ – \(\frac{2 x}{y}\) + log |x| = C …………….(1)
When x = e, y = e
∴ from (1) ;
– \(\frac{2 e}{e}\) + log |e| = C
⇒ – 2 + 1 = C
⇒ C = – 1
Thus, eqn. (1) becomes ;
– \(\frac{2 x}{y}\) + log |x| = – 1
⇒ 1 + log |x| = \(\frac{2 x}{y}\)
⇒ y = \(\frac{2 x}{1+\log |x|}\) be the required particular solution.

(iii) Given, (xy – y²) dx – x² dy = 0
⇒ \(\frac{d y}{d x}=\frac{x y-y^2}{x^2}\) …………..(1)
Also, \(\frac{d y}{d x}=\frac{y}{x}-\left(\frac{y}{x}\right)^2=\phi\left(\frac{y}{x}\right)\)
Thus, eqn. (1) be a homogeneous differential eqn.
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ from (1) ;

(iv) Given diff. eqn. can be written as ;

(v) Given \(\sqrt{1-y^2}\) dx = (sin^-1 y – x) dy
⇒ \(\frac{d x}{d y}+\frac{x}{\sqrt{1-y^2}}=\frac{\sin ^{-1} y}{\sqrt{1-y^2}}\)
which is L.D.E. in x and is of the form

⇒ xe^{sin-1 y} = ∫ t e^t dt + C
= (t – 1) e^t + C
⇒ xe^{sin-1 y} = (sin^-1 y – 1) e^{sin-1 y} + C ……………..(1)
When x = 0, y = 0
∴ from (1) ;
0 = (0 – 1) + C
⇒ C = 1
∴ from (1) ; we get
xe^{sin-1 y} = (sin^-1 y – 1) e^{sin-1 y} + 1
⇒ x = sin^-1 y – 1 + e^{– sin-1 y}
which gives the required particular solution.

(vi) Given diff. eqn. can be written as

∴ I = – sin v e^-v + [- cos v e^-v – ∫ sin v e^-v dv]
⇒ 2I = – (sin v + cos v) e^-v
∴ from (2) ; we get
\(\frac{-(\sin v+\cos v) e^{-v}}{2}\) = – log |x| + C
⇒ e^-y/x (sin \(\frac{y}{x}\) + cos \(\frac{y}{x}\)) = 2 log |x| + A …………….(3)
where A = – 2C
When x = 1, y = 0
∴ from (3) ; we have
1 (0 + 1) = 2 × 0 + A
⇒ A = 1
Thus eqn. (4) becomes ;
e^-y/x (sin \(\frac{y}{x}\) cos \(\frac{y}{x}\)) = 2 log |x| + 1
which gives the required particular solution of given diff. equation.

The availability of step-by-step ML Aggarwal Class 12 Solutions ISC Chapter 9 Differential Equations Ex 9.8 can make challenging problems more manageable.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.8

Question 1.
The surface area of a balloon being inflated changes at a constant rate. If initially, its radius is 3 units and after 2 seconds, it is 5 units, find the radius after t seconds.
Solution:
Let r be the radius of bal loon.
Then surface area of balloon = 4πr²
according to given condition, we have
\(\frac{d}{d t}\) (4πr²) = K (say)
⇒ r dr = \(\frac{K}{8 \pi}\) dt ;
on integrating
⇒ \(\frac{r^2}{2}=\frac{K}{8 \pi} t+\mathrm{C}\) ………………….(1)
initially t = 0 ; r = 3 units
∴ from (1) ;
\(\frac{9}{2}\) = C
Thus eqn. (1) gives ;
\(\frac{r^2}{2}=\frac{K}{8 \pi} t+\frac{9}{2}\)
⇒ r² = \(\frac{K}{4 \pi}\) + 9 ………………..(2)
Also when t = 2 seconds, r = 5 units
∴ from (2) ; we have
25 = \(\frac{K}{4 \pi}\) × 2 + 9
⇒ 16 = \(\frac{K}{2 \pi}\)
⇒ K = 32 π
∴ from (2) ;
r² = \(\frac{32 \pi}{4 \pi}\) t + 9
= 8t + 9
⇒ r = \(\sqrt{8 t+9}\)
which is required radius after time t.

Question 2.
The surface area of a balloon being inflated changes at a rate proportional to time. If initially its radius is 1 unit and after I second ¡t is 3 units, find its radius after time t.
Solution:
Let r be the radius after time t
Then \(\frac{d}{d t}\) (4πr²) ∝ t
⇒ \(\frac{d}{d t}\) (4πr²) = Kt
where K be the constant of proportionality
8πr \(\frac{d r}{d t}\) = Kt
⇒ r dr = \(\frac{K}{8 \pi}\) t dt
On integrating ; we have
\(\frac{r^2}{2}=\frac{K}{8 \pi} \times \frac{t^2}{2}+\mathrm{C}\) …………….(1)
initially t = 0, r = 1 units
∴ from (1) ;
\(\frac{1}{2}=\frac{\mathrm{K}}{16 \pi}\) × 0 + C
⇒ C = \(\frac{1}{2}\)
∴ from (1) ; we have
\(\frac{r^2}{2}=\frac{K}{16 \pi} t^2+\frac{1}{2}\) …………………….(2)
\(r^2=\frac{\mathrm{K}}{8 \pi} t^2+1\)
When t = 1 second ;
r = 3 units
∴ from (2) ;
9 = \(\frac{K}{8 \pi}\) + 1
K = 64 π
∴ from (2) ;
r² = \(\frac{64 \pi}{8 \pi}\) t² + 1
r = \(\sqrt{8 t^2+1}\) units
which is the required radius after time t.

Question 3.
A population grows at the rate of 2 % per year. How long does it take for the population to double itself?
Solution:
Let P₀ be the population initially i.e. at t = 0
and P be the population after t years.
According to given condition, we have
\(\frac{d P}{d t}\) = 2% of P
\(\frac{d P}{d t}\) = \(\frac{2}{100}\) P
= \(\frac{t}{50}\) P
On integrating ; we have
log P = \(\frac{t}{50}\) + C ………………(1)
When t = 0; P = P₀
∴ from (1) ;
log P₀ = C
Thus eqn. (1) gives;
log P = \(\frac{t}{50}\) + log P₀
⇒ log \(\frac{\mathrm{P}}{\mathrm{P}_0}=\frac{t}{50}\) ……………..(2)
Let t = t₁
When P = 2P₀
∴ from (2) ;
log \(\frac{2 \mathrm{P}_0}{\mathrm{P}_0}=\frac{t_1}{50}\)
⇒ t₁ = 50 log 2
Hence it takes 50 log 2 years for the population to grow 2 times.

Question 4.
It is given that the rate at which some bacteria multiply is proportional to the instantaneous number present. If the original number of bacteria doubles in 2 hours, in how many hours will it be five times ?
Solution:
Let P₀ be the number of bacteria present initially
i.e. t = 0 and P be the bacteria present after t years.
Then \(\frac{d P}{d t}\) ∝ P
⇒ \(\frac{d P}{d t}\) = KP ;
where K = constant of proportionality
on integrating ; we have
∫ \(\frac{d P}{P}\) = K ∫ dt
⇒ log P = Kt + C …………….(1)
When t = 0 ; P = P₀
∴ from (1) ;
log P₀ = C
Thus eqn. (1) gives ;
log \(\frac{\mathrm{P}}{\mathrm{P}_0}\) = Kt ……………..(2)
also P = 2P₀ ; t = 2 hours
∴ from (2);
log \(\frac{2 \mathrm{P}_0}{\mathrm{P}_0}\) = 2K
⇒ K = \(\frac{1}{2}\) log 2
Thus eqn. (2) gives;
log \(\frac{\mathrm{P}}{\mathrm{P}_0}=\left(\frac{1}{2} \log 2\right) t\) ………………(3)
Let t = t₁ when P = 5P₀
∴ eqn. (3) gives ;
log \(\frac{5 \mathrm{P}_0}{\mathrm{P}_0}\) = (\(\frac{1}{2}\) log 2) t₁
⇒ t₁ = \(\frac{2 \log 5}{\log 2}\) hours
Hence \(\frac{2 \log 5}{\log 2}\) hours, bacteria will be five times the no. of bacteria present initially.

Question 5.
Assume that a spherical rain drop evaporates at a rate proportional to its surface area. If its radius originally is 3 mm and 1 hour later has been reduced to 2 mm, find an expression for the radius of the rain drop at any time.
Solution:
Let r be the radius of rain drop after timer.
Then according to given condition,
\(\frac{d}{d t}\left(\frac{4}{3} \pi r^3\right)\) ∝ 4πr²
⇒ \(\frac{d}{d t}\left(\frac{4}{3} \pi r^3\right)\) = – K 4πr²
where K be the constant of proportionality
\(\frac{4 \pi}{3} \times 3 r^2 \frac{d r}{d t}\) = – K 4πr² ;
⇒ \(\frac{d r}{d t}\) = – K
on integrating
⇒ dr = – K ∫ dt + C
⇒ r = – Kt + C …………..(1)
When t = 0 : r = 3
∴ from (1) ;
3 = C
∴ eqn. (1) gives;
r = – Kt + 3 ………………(2)
further when t= 1 hour, r = 2 mm
∴ from (2) ;
2 = – K + 3
⇒ K = 1
putting the value of K in eqn. (2); we get
r = – t + 3 ; 0 < t < 3
Which gives the required radius of rain drop atany time t.

Question 5 (old).
A wet porous substance in the open air loses its moisture at a rate proportional to the moisture content. If a sheet hung in the mind loses half its moisture the first hour, when will it have lost 90% moisture, whether conditions remaining the same ?
Solution:
Let M₀ be the moisture content initially at t = 0
and M be the moisture constant after time t hours.
Then according to given condition, we have
\(\frac{d \mathrm{M}}{d t}\) ∝ M
\(\frac{d \mathrm{M}}{d t}\) = – KM
where K = constant of proportionality
\(\frac{d \mathrm{M}}{\mathrm{M}}\) = – K dt [after variable separation]
∫ \(\frac{d \mathrm{M}}{\mathrm{M}}\) = ∫ – K dt + C [on integrating]
⇒ log M = – Kt + C …………………..(1)
When t = 0 ;
M = M₀
∴ from (1) ;
log M₀ = C
∴ eqn. (1) gives ;
log M = – Kt + log M₀
⇒ log \(\frac{\mathrm{M}}{\mathrm{M}_0}\) = – Kt ………………..(2)
When t = 1, M = \(\frac{1}{2}\) M₀
∴ from (1) ;
log \(\left(\frac{\frac{1}{2} M_0}{M_0}\right)\) = – K × 1
⇒ – K = log \(\frac{1}{2}\)
⇒ – K = log 1 – log 2
⇒ – K = – log 2
⇒ K = log 2
putting the value of K in eqn. (2) ; we have
log \(\left(\frac{M}{M_0}\right)\) = – (log 2) t ………………….(3)
When the sheet loses 90% of the moisture.
i.e. When M = M₀ – \(\frac{90}{100}\) M₀
= \(\frac{\mathrm{M}_0}{10}\) t = t₁
∴ from (3) ;
log \(\left(\frac{\mathrm{M}_0}{10 \mathrm{M}_0}\right)\) = – (log 2) t₁
⇒ t₁ = \(\frac{1}{-\log 2} \log \frac{1}{10}\)
= \(\frac{\log 1-\log 10}{-\log 2}\)
= \(\frac{\log 10}{\log 2}\)

Question 6.
The acceleration of a particle moving along the line OX at any time t seconds is given by a = 2 (t – 5). If at t = 0, the velocity and displacement of the particle are 9 m/sec and 18 m respectively, find its velocity v and displacement x at any time t seconds. What is the velocity at t = 1 sec?
Solution:
Given acceleration of particle at any time t is given by
a = \(\frac{d^2 x}{d t^2}\)
= 2 (t – 5)
= \(\frac{d v}{d t}\) ……………….(1)
where v be the velocity of particle at any time t.
On integrating eqn. (1) ; we have
v = 2 \(\left(\frac{t^2}{2}-5 t\right)\) + C ………………(2)
given at t = 0 ; v = 9 m/sec
∴ from (2) ;
9 = 0 + C
⇒ C = 9
Substituting the value of C in eqn. (2) ; we have
\(\frac{d x}{d t}\) = v = t² – 10t + 9 ………………(3)
where x be the displacement of particle at any time t.
On integrating both sides of eqn. (3) ; we have
x = \(\frac{t^3}{3}-10 \frac{t^2}{2}+9 t+C_1\) ………………(4)
Further at t = 0; x = 18 m
∴ from (4) ;
18 = C₁
Thus, eqn. (4) gives;
x = \(\frac{t^3}{3}\) – 5t² + 9t + 18 ………………(5)
From (3) ;
(v)_{t = 1} = 1 – 10 + 9 = 0 m/sec.

Question 7.
Find the equation of the curve through the point (1, 0) if the slope of the tangent to the curve at any point (x, y) is \(\frac{y-1}{x^2+x}\). (NCERT Exemplar)
Solution:
Given slope of tangent to the curve at any point (x, y) = \(\frac{y-1}{x^2+x}\)
Also, slope of tangent to curve at any point (x, y) = \(\frac{d y}{d x}\)

which gives the family of curves satisfied by diff. eqn. (1).
Now curve given by eqn. (2) passes through (1, 0).
i.e. when x = 1, y = 0
∴ from (2) ;
– 1 = \(\frac{C}{2}\)
⇒ C = – 2
putting the value of C in eqn. (2); we have
(y – 1) (x + 1) + 2x = 0, which is the required eqn. of curve.

Question 8.
Find the equation of a curve passing through the point (2, 1) if the slope of the tangent to the curve at any point (x, y) is \(\frac{x^2+y^2}{2 x y}\).
Solution:
We know that, slope of tangent at any point (x, y) to the curve = \(\frac{d y}{d x}\)
Also, slope of tangent to curve at any point (x, y) = \(\frac{x^2+y^2}{2 x y}\)
∴ \(\frac{d y}{d x}=\frac{x^2+y^2}{2 x y}\) ………………..(1)
which is homogeneous differential equation
Put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)

which represents the family of curves satisfied by given differential equation (1).
Since the particular member of the family of curve passes through the point (2, 1);
we have
1² – 2² = C × 2
⇒ 2C = – 3
⇒ C = – \(\frac{3}{2}\)
putting the value of C in eqn. (2); we get
(y² – x²) = – \(\frac{3 x}{2}\)
⇒ 2 (x² – y²) = 3x,
which is the required equation of curve.

Question 8 (old).
Find the equation of a curve passing through the point (1, 1) whose differential equation ¡s x dy = (2x² + 1) dx, x ≠ 0. (NCERT)
Solution:
Given differential eqn. can be written as
\(\frac{d y}{d x}=\frac{2 x^2+1}{x}\) ;
on integrating
y = ∫ \(\frac{2 x^2+1}{x}\) dx + C
⇒ y = \(\frac{2 x^2}{2}\) + log |x| + C …………….(1)
Now (1) represents family of curves.
Now the curve passes through (I, 1)
∴ from (1) ;
1 = 1 + C
⇒ C = 0
∴ eqn.(1) gives ;
y = x² + log |x| istherequired particular solution.

Question 9.
Find the equation of a curve passing through the origin given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the coordinates of the point. (NCERT)
Solution:
We know that slope of tangent at any poiont (x, y) of the curve = \(\frac{d y}{d x}\)
also according to given condition, we have
\(\frac{d y}{d x}\) = x + y
⇒ \(\frac{d y}{d x}\) – y = x
which is linear differential equation of the form
\(\frac{d y}{d x}\) + Py = Q ;
where P = – 1 and Q = x
∴ I.F. = e^{∫ P dx}
= e^{∫ – 1 dx}
= e^{– x}
and solution is given by
ye^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + c
⇒ ye^{– x} = ∫ x e^{– x} dx + c
⇒ y e^{– x} = [- x e^{– x} – e^{– x}] + c
⇒ y = – (x + 1) + ce^x ……………………(1)
it is given that curve given by eqn (1) passes through (0, 0)
i.e. when x = 0 ; y = 0
∴ from (1) ; we have
0 = – (0 + 1) + c
⇒ c = 1
Thus from eqn. (1) ; we have
y = – x – 1 + e^x
⇒ x + y = e^x – 1
which gives the required equation of curve.

Question 10.
Find the equation of a curve passing through the point (0, 2) given that the sum of the coordinates of any point (x, y) on the curve exceeds the slope of the tangent to the curve at the point by 5. (NCERT)
Solution:
We know that,
slope of tangent to curve at any point (x,y) = \(\frac{d y}{d x}\)
according to given condition, we have
\(\frac{d y}{d x}\) + 5 = x + y
⇒ \(\frac{d y}{d x}\) – y = x – 5
which is linear duff. eqn. in y and is of the form \(\frac{d y}{d x}\) + Py = Q
where P = – 1 and Q = x – 5
⇒ IF. = e^{∫ P dx}
= e^{∫ – dx}
= e^{– x}
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ ye^-x = ∫ (x – 5) e^{– x} dx + C
⇒ ye^{– x} = (x – 5) e^{– x} – e^{– x} + C
⇒ ye^{– x} = – (x – 4) e^{– x} + C
⇒ y = – (x – 4) + C e^{– x} …………….(1)
Since the curve passing through the point (0, 2)
∴ eqn. (1) passes through (0, 2).
i.e. when x = 0 ; y = 2
∴ from (1) ;
2 = – (0 – 4) + C
⇒ C = – 2
Thus eqn. (1) gives ;
y = – (x – 4) – 2 e^x
which gives the required eqn. of curve.

Question 10 (old).
Find the equation of a curve passing through the point (- 2, 3) given that the slope of the tangent to the curve at any point (x, y) is \(\frac{2 x}{y^2}\). (NCERT)
Solution:
Given slope of tangent to curve at any point (x, y) = \(\frac{2 x}{y^2}\)
Also, slope of tangent to curve at any point (x, y) = \(\frac{d y}{d x}\) ……………….(1)
∴ \(\frac{d y}{d x}=\frac{2 x}{y^2}\)
⇒ ∫ y² dy = ∫ 2x dx + C
On integrating; we have
∫ y² dy = ∫ 2x dx + c
⇒ \(\frac{y^3}{3}\) = x² + C ………………….(2)
So eqn. (2) represents the family of curves satisfied by diff eqn. (1).
Now given curve (2) passes through (- 2, 3),
i.e., when x = – 2, y = 3
∴ from (2) ;
9 = 4 + C
⇒ C = 5
Thus, from (2) gives ;
⇒ \(\frac{y^3}{3}\) = x² + 5
⇒ y³ = 3x² + 15, which is the required equation of curve.

Students often turn to ML Aggarwal Class 12 Solutions Chapter 9 Differential Equations Ex 9.7 to clarify doubts and improve problem-solving skills.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.7

short answer type questions (1 to 5):

Question 1.
Which of the following equations are linear differential equations of first order in y?
(i) \(\frac{d y}{d x}\) + 3y = sin x
(ii) x² \(\frac{d y}{d x}\) – 2xy = x³ + 2x² – 5
(iii) x log x dy + (3y – 5 log x) dx = 0
(iv) y \(\frac{d y}{d x}\) + 5x = y cos x.
Solution:
(i) Given diff. eqn. be,
\(\frac{d y}{d x}\) + 3y = sin x
which is of the form, \(\frac{d y}{d x}\) + Py = Q
where P = 3 and Q = sin x
and both are functions of x-alone.
Thus given diff. eqn. is linear diff. eqn. in y of 1st order.

(ii) Given diff. eqn. can be written as
\(\frac{d y}{d x}-\frac{2}{x} y=\frac{x^3+2 x^2-5}{x^2}\) ……………….(1)
which is o fthe form \(\frac{d y}{d x}\) + Py = Q
where P = – \(\frac{2}{x}\) and Q = \(\frac{x^3+2 x^2-5}{x^2}\)
i.e. P and Q are functions of x-alone.
Hence eqn. (1) is L.D.E in y of first order.

(iii) Given diff. eqn. can be written as,
x log x dy + (3y – 5 log x) dx = 0
⇒ \(\frac{d y}{d x}+\left(\frac{3}{x \log x}\right) y=\frac{5}{x}\) ………………….(1)
which is of form \(\frac{d y}{d x}\) + Py = Q;
where P = \(\frac{3}{x \log x}\) and Q = \(\frac{5}{x}\);
and both are functions of x-alone.
Thus eqn. (1) be a L.D.E my of first order.

(iv) Given diff. eqn. can be written as,
\(\frac{d y}{d x}+\frac{5}{y} x\) = cos x
which is not of the form \(\frac{d y}{d x}\) + Py = Q
where P and Q are functions of x-alone.
Thus eqn. (1) is not a L.D.E in y of first order.

Question 2.
Which of the following equations are linear differential equations of first order in x?
(i) \(\frac{d x}{d y}-\frac{2 x}{y}\) = 3y³ – 5y + 1
(ii) (cos y) \(\frac{d x}{d y}\) – 2x sin y = cos y
(iii) (1 + y²) dx = (tan^-1 y – 2x) dy
(iv) x² \(\frac{d x}{d y}\) + 2x³y = 3y² – 2.
Solution:
(i) Given diff. eqn. be,
\(\frac{d x}{d y}-\frac{2 x}{y}\) = 3y³ – 5y + 1
which is ofthe form \(\frac{d x}{d y}\) + Px = Q
where P = – \(\frac{2}{y}\) ; Q = 3y³ – 5y + 1
both are functions of y-alone.
Thus, given diff. eqn. be a L.D.E in x of first order.

(ii) Given diff. eqn. can be written as,
\(\frac{d y}{d x}+\frac{1}{2 x} y=\frac{6 x^3+5}{2 x}\)
which is ofthe form \(\frac{d x}{d y}\) + Px = Q
where P = \(\frac{1}{2 x}\)
and Q = \(\frac{6 x^3+5}{2 x}\)
∴ I.F . = e^{∫ P dx}
= e^{\(\frac{1}{2 x}\)}
= e^{\(\frac{1}{2}\) log |x|}
= \(e^{\log x^{\frac{1}{2}}}\) = √x

(iii) Given diff. eqn. be.
\(\frac{d y}{d x}\) + 2y tan x = sin x.
which is L.D.E in y
and which is of the form \(\frac{d y}{d x}\) + Py = Q
Here P = 2 tan x; Q = sin x
∴ I.F. = e^{∫ P dx}
= e^{∫ 2 tan x dx}
= \(e^{-2 \int \frac{-\sin x d x}{\cos x}}\)
= e^{– 2 log |cos x|}
= e^{log |cos x\-2}
= \(\frac{1}{\cos ^2 x}\)
= sec² x

(iv) Given diff eqn. can be written as,
\(\frac{d y}{d x}+\frac{y}{x}\) = y cot x
⇒ \(\frac{d y}{d x}\) + (\(\frac{1}{x}\) – cot x) y = 0
which is L.D.E in y
Here P = \(\frac{1}{x}\) – cot x; Q = 0
∴ I.F. = e^{∫ P dx}
= \(e^{\int\left(\frac{1}{x}-\cot x\right) d x}\)
= e^{log x – log sin x}
= \(e^{\log \frac{x}{\sin x}}\)
= \(\frac{x}{\sin x}\)
= x cosec x

Question 4.
Write an integrating factor of each of the following linear differential equation:
(i) y dx – (x + 2y²) dy = 0, y > 0
(ii) y e^y dx = (y³ + 2x e^y) dy, y ≠ 0
(iii) dx + x dy = (e^{– y} log y) dy, y > 0
(iv) (1 – y²) \(\frac{d x}{d y}\) + yx = ay, |y| < 1 (NCERT)
Solution:
(i) Given diff. eqn. can he written as,
\(\frac{d x}{d y}-\frac{x}{y}\) = 2y,
which is L.D.E in x and is of the form \(\frac{d x}{d y}\) + Px = Q
where P = – \(\frac{1}{y}\) and Q = 2y
∴ I.F. = \(e^{\int-\frac{1}{y} d y}\)
= e^{– log y}
= e^{log y-1}
= \(\frac{1}{y}\)

(ii) Given diff. eqn. can be written as,
\(\frac{d x}{d y}=\frac{y^3+2 x e^y}{y e^y}\)
⇒ \(\frac{d x}{d y}-\frac{2}{y} x=\frac{y^2}{e^y}\)
which is L.D.E in x and is of the form
\(\frac{d x}{d y}\) + Px = Q,
where P \(\frac{2}{y}\)
and Q = \(\frac{y^2}{e^y}\)
∴ I.F = e^{∫ P dy}
= \(e^{\int-\frac{2}{y} d y}\)
= e^{– 2 log y}
= e^{log y-2}
= \(\frac{1}{y^2}\)

(iii) Given diff. eqn. can be written as,
dx = (e^{– y} log y – x) dy
⇒ \(\frac{d x}{d y}\) + Px = Q
which is L.D.E in x and is of the form
\(\frac{d x}{d y}\) + Px = Q
where P = \(\frac{y}{1-y^2}\)
and Q = \(\frac{a y}{1-y^2}\)
I.F. = e^{∫ P dy}
= \(e^{\int \frac{y}{1-y^2} d y}\)
= \(e^{-\frac{1}{2} \int \frac{-2 y d y}{1-y^2}}\)
= \(e^{-\frac{1}{2} \log \left(1-y^2\right)}\)
= e^{log (1 – y2)– 1/2}
= (1 – y²)^-1/2
= \(\frac{1}{\sqrt{1-y^2}}\)

Question 5.
(i) \(\frac{d y}{d x}+\frac{y}{x}\) = x² (NCERT)
(ii) x \(\frac{d y}{d x}\) + 2y = x²
Solution:
(i) Given eqn. be \(\frac{d y}{d x}+\frac{y}{x}\) = x²,
it is L.D.E of Ist order
On comparing with \(\frac{d y}{d x}\) + Py = Q ; we get
P = \(\frac{1}{x}\) ; Q = x²
I.F. = e^{∫ P dx}
= e^{∫ \(\frac{1}{x}\) dx}
= e^{log x} = x
Multiply given eqn. by x ; we get
x \(\frac{d y}{d x}\) + y = x³
⇒ \(\frac{d}{d x}\) (xy) = x³
on integrating; we have
xy = ∫ x³ dx + C
⇒ xy = \(\frac{x^4}{4}\) + c
⇒ y = \(\frac{x^3}{4}+\frac{C}{x}\) be the required solution.

(ii) Given diff. eqn. can be written as
\(\frac{d y}{d x}+\frac{2}{x} y\) = x
which is L.D.E. in y of first order and is of the form
\(\frac{d y}{d x}\) + Py = Q.
Here P = \(\frac{2}{x}\) ; Q = x
∴ I.F. = e^{∫ P dx}
= \(\int_e \frac{2}{x} d x=e^{2 \log x}\)
= e^{log x2}
= x²
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ y . x² = ∫ x . x² dx + C
⇒ x²y = \(\frac{x^4}{4}\) + C
⇒ y = \(\frac{x^2}{4}\) + Cx^-2
which is the required solution.

Question 5 (old).
Find the general solution of the differential equation e^x dy + (y e^x + 2x) dx = 0. (NCERT)
Solution:
Given differential eqn. can be written as,
e^x dy + (y e^x + 2x) dx = 0
⇒ e^x \(\frac{d y}{d x}\) + y e^x + 2x = 0
⇒ \(\frac{d y}{d x}\) + y = – \(\frac{2 x}{e^x}\)
which is LD.E in y and is of the form
\(\frac{d y}{d x}\) + Py = Q
Here P = 1 and Q = – \(\frac{2 x}{e^x}\)
∴ I.F. = e^{∫ P dx}
= e^{∫ 1 . dx}
= e^x
and solution is given by
y . e^{∫ P dx} = ∫ (Q . e^{∫ P dx}) dx + C
⇒ y . e^x = ∫ – \(\frac{2 x}{e^x}\) . e^x dx + C
⇒ y e^x = – x² + C, which gives the required solution.

Question 6.
(i) \(\frac{d y}{d x}\) + 2y = 6e^x
(ii) \(\frac{d y}{d x}\) + 3y = e^{– 2x} (NCERT)
Solution:
(i) Given eqn. is \(\frac{d y}{d x}\) + 2Y = 6e^x …………….(1)
Now eqn. (1) is LD.E of 1st order in y.
On comparing with \(\frac{d y}{d x}\) + Py = Q;
we have
P = 2 and Q = 6e^x
∴ I.F = e^{∫ P dx}
= e^{∫ 2 dx}
= e^2x
Multiply eqn. (1) by e^2x; we get
e^2x \(\frac{d y}{d x}\) + 2 e^2x y = 6 e^3x
⇒ \(\frac{d}{d x}\) (y e^2x) = 6 e^3x ;
On integrating we have
y e^2x = ∫ 6 e^3x dx + C
⇒ y e^2x = \(\frac{6 e^{3 x}}{3}\) + C
⇒ y = 2 e^x + C e^-2x is the required solution.

(ii) Given eqn. is \(\frac{d y}{d x}\) + 3y = e^{– 2x}, which ¡s L.D.E of 1st order.
On comparing with \(\frac{d y}{d x}\) + Py = Q, we have
P = 3 and Q = e^{– 2x}
and I.F. = e^{∫ P dx}
= e^{∫ 3 dx}
= e^3x
Multiply given eqn. by e^3x, we get
e^3x \(\frac{d y}{d x}\) + 3 e^3x y = e^{– 2x} . e^3x
⇒ \(\frac{d}{d x}\) (e^3x . y) = e^x
On integrating ; we have
e^3x . y = ∫ e^x dx + C
⇒ e^3x y = e^x + C
⇒ y = e^{– 2x} + C e^{– 3x}

Question 7.
(i) (1 + x²) \(\frac{d y}{d x}\) = 4x² – 2xy (ISC 2020)
(ii) \(\frac{d y}{d x}+\frac{y}{x}\) = e^{– x}
Solution:
(i) Given differential equation can be written as ;

(ii) Given \(\frac{d y}{d x}+\frac{y}{x}\) = e^{– x}
which is L.D.E. in y and is of the form
\(\frac{d y}{d x}\) + Py = Q,
where P = \(\frac{1}{x}\) and Q = e^{– x}
∴ I.F. = e^{∫ P dx}
= e^{∫ \(\frac{1}{x}\) dx}
= e^{log x} = x
∴ I.F. = e^{∫ P dx}
= ∫ Q . e^{∫ P dx} dx + C
⇒ yx = ∫ e^{– x} . x dx + C
xy = x \(\frac{e^{-x}}{-1}\) + ∫ 1 . e^{– x} dx + C
⇒ xy = – x e^{– x} – e^{– x} + C
⇒ y = \(-\left(1+\frac{1}{\dot{x}}\right) e^{-x}+\frac{\mathrm{C}}{x}\)
which gives the required solution.

Question 8.
(i) (x² + 1) \(\frac{d y}{d x}\) + 2xy = \(\sqrt{x^2+4}\)
(ii) x \(\frac{d y}{d x}\) + y = 3x cos 2x , x > 0
Solution:
(i) Given, (x² + 1) \(\frac{d y}{d x}\) + 2xy = \(\sqrt{x^2+4}\)

(ii) Given, x \(\frac{d y}{d x}\) + y = 3x cos 2x

Question 9.
(i) x \(\frac{d y}{d x}\) – y = (x – 1) e^x, x > 0
(ii) \(\frac{d y}{d x}\) + x = tan y + sec² y
Solution:
(i) Given diff. eqn. be,
x \(\frac{d y}{d x}\) – y = (x – 1) e^x
i.e., \(\frac{d y}{d x}-\frac{y}{x}=\left(\frac{x-1}{x}\right) e^x\)
which is L.D.E of the form \(\frac{d y}{d x}\) + Py = Q
where P = – \(\frac{1}{x}\)
and Q = \(\frac{(x-1)}{x}\) e^x

(ii) Given diff. eqn. be,
\(\frac{d y}{d x}\) + x = tan y + sec² y
which is linear diff. eqn. in x and is of the form \(\frac{d y}{d x}\) + Px = Q
where P = 1 ; Q = tan y + sec² y
∴ I.F. = e^{∫ P dy}
= e^{∫ dy} = e^y
⇒ xe^y = ∫ (tan y + sec² y) e^y dy + C
= ∫ e^y tan y dy + ∫ e^y sec² y dy + C
= tan y e^y – ∫ sec² y e^y dy + ∫ e^y sec² y dy + C
⇒ x e^y = tan y e^y + C
⇒ x = tan y + C e^-y

Question 10.
(i) x \(\frac{d y}{d x}\) + 2y = x cos x
(ii) (sin x) \(\frac{d y}{d x}\) + (cos x) y = cos x sin² x
(iii) sin x \(\frac{d y}{d x}\) – y = sin x tan \(\frac{x}{2}\)
Solution:
(i) Given diff. eqn. be,
x \(\frac{d y}{d x}\) + 2y = x cos x
⇒ \(\frac{d y}{d x}\) + \(\frac{2}{x}\) y = cos x
which is L.D.E. of the form \(\frac{d y}{d x}\) + Px = Q
where P = \(\frac{2}{x}\) and Q = cos x
Now I.F. = e^{∫ P dx}
= \(e^{\int \frac{2}{x} d x}\)
= \(e^{2 \int \frac{1}{x} d x}\)
= e^{2 log |x|}
= x²
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} + c
⇒ y . x² = ∫ cos x . x² dx + c
= x² sin x – ∫ 2x sin x dx + c
⇒ yx² = x² sin x – 2 [- x cos x + sin x] + c
⇒ x²y = x² sin x + 2x cos x – 2 sin x + c be the required solution.

(ii) Given eqn. be
sin x \(\frac{d y}{d x}\) + cos x . y = cos x . sin² x
⇒ \(\frac{d y}{d x}\) + (\(\frac{\cos x}{\sin x}\)) y = cos x . sin x …………………(1)
On comparing (1) with \(\frac{d y}{d x}\) + Px = Q
P = \(\frac{\cos x}{\sin x}\) ;
Q = cos x sin x
∴I.F. = \(e^{\int \frac{\cos x d x}{\sin x}}\)
= e^{log sin x}
= sin x
Multiply eqn. (1) by sin x ; we get
sin x \(\frac{d y}{d x}\) + y cos x = cos x sin² x
⇒ \(\frac{d}{d x}\) (y sin x) = cos x . sin² x ;
On integrating ; we get
y sin x = ∫ (sin x)² cos x dx + C
= \(\frac{(\sin x)^3}{3}\) + C
⇒ y = \(\frac{\sin ^2 x}{3}+\frac{C}{\sin x}\) is the required solution.

(iii) Given diff. eqn. be,
sin x \(\frac{d y}{d x}\) – y = sin x tan \(\frac{x}{2}\)
⇒ \(\frac{d y}{d x}\) – y cosec x = tan \(\frac{x}{2}\)
On comparing with \(\frac{d y}{d x}\) + Py = Q
where P = – cosec x ;
Q = tan \(\frac{x}{2}\)
∴I.F. = e^{∫ P dx}
= e^{– ∫ cosec x dx}
= \(e^{-\log \left|\tan \frac{x}{2}\right|}\)
= cot \(\frac{x}{2}\)
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ y . cot \(\frac{x}{2}\) = ∫ tan \(\frac{x}{2}\) . cot \(\frac{x}{2}\) + C
⇒ y . cot \(\frac{x}{2}\) = x + C, which is the required solution.

Question 11.
(i) \(\frac{d y}{d x}=-\frac{x+y \cos x}{1+\sin x}\)
(ii) \(\frac{d y}{d x}\) + y cos x = e^{sin x} cos x
Solution:
(i) Given, \(\frac{d y}{d x}=-\frac{x+y \cos x}{1+\sin x}\)
⇒ \(\frac{d y}{d x}+\left(\frac{\cos x}{1+\sin x}\right) y=\frac{-x}{1+\sin x}\)
which is L.D.E. in y.
Here P = \(\frac{\cos x}{1+\sin x}\) ;
Q = \(\frac{-x}{1+\sin x}\)
∴ I.F. = e^{∫ P dx}
= \(e^{\int \frac{\cos x d x}{1+\sin x}}\)
= e^{log (1 + sin x)}
= 1 + sin x
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ y (1 + sin x) = ∫ \(\frac{-x}{1+\sin x}\) × (1 + sin x) dx + C
⇒ y (1 + sin x) = – \(\frac{x^2}{2}\) + C,
which is the required solution.

(ii) Given diff. eqn. be,
\(\frac{d y}{d x}\) + y cos x = e^{sin x} cos x

where P = cos x and Q = e^{sin x} cos x
Now I.F. = e^{∫ P dx}
= e^{∫ cos x dx}
= e^{sin x}
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + c
⇒ y . e^{sin x} = e^{sin x} cos x . e^{sin x} dx + c
= ∫ e^{2 sin x} cos x dx + c
put sin x = t
⇒ cos x dx = dt
y e^{sin x} = ∫ e^{2 t} dt + c
⇒ y e^{sin x} = \(\frac{e^{2 t}}{2}\) + c
⇒ y e^{sin x} = \(\frac{1}{2}\) e^{2 sin x} + c
⇒ y = \(\frac{1}{2}\) e^{sin x} + ce^{– sin x}
which is the required solution curve.

Question 12.
(i) \(\frac{d y}{d x}\) – tan x . y = e^{sin x}, 0 < x < \(\frac{\pi}{2}\)
(ii) \(\frac{d y}{d x}\) + y = cos x – sin x
Solution:
(i) Given,
\(\frac{d y}{d x}\) – tan x . y = e^{sin x} ; 0 < x < \(\frac{\pi}{2}\)
which is linear diff. eqn. in y
Here P = – tan x and Q = e^{sin x}
∴ I.F. = e^{∫ P dx}
= e^{∫ – tan x dx}
= \(e^{\int \frac{-\sin x}{\cos x}} d x\)
= e^{∫ log cos x dx}
= cos x
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ y . cos x = ∫ e^{sin x} cos x dx + C
put sin x = t
⇒ cos x dx = dt
y cos x = ∫ e^t dt + C
⇒ y cos x = e^{sin x} + C
⇒ y = sec x (e^{sin x} + C).

(ii) Given diff. eqn. be,
\(\frac{d y}{d x}\) + y = cos x – sin x
which is L.D.E. in y of first order and on
comparing with \(\frac{d y}{d x}\) + Py = Q ; we have
P = 1 and Q = cos x – sin x
∴ I.F. = e^{∫ P dx}
= e^{∫ 1 dx}
= e^x
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} + C
ye^x = ∫ e^x (cos x – sin x) + C
= ∫ e^x cos x dx – ∫ e^x sin x dx + C
= cos x e^x + ∫ sin x e^x dx – ∫ e^x sin x dx + C
⇒ ye^x = cos x e^x + C
⇒ y = cos x + C e^{– x}
which is the required solution.

Question 13.
(i) x \(\frac{d y}{d x}\) + y = x log x
(ii) x \(\frac{d y}{d x}\) + 2y = x² log x. (NCERT)
Solution:
(i) Given diff. eqn. be,
x \(\frac{d y}{d x}\) + y = x log x
⇒ \(\frac{d y}{d x}+\frac{y}{x}\) = log x
which is L.D.E. of the form \(\frac{d y}{d x}\) + Py = Q
where P = \(\frac{1}{x}\)
and Q = log x
Now I.F. = e^{∫ P dx}
= \(e^{\int \frac{1}{x} d x}\)
= e^{log x} = x
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} + c
⇒ y.x = ∫ log x . x dx + c
⇒ xy = (log x) \(\frac{x^2}{2}\) – \(\int \frac{1}{x} \cdot \frac{x^2}{2} d x\) + c
⇒ xy = \(\frac{x^2}{2}\) log x – \(\frac{x^2}{4}\) + c
⇒ 4xy = 2x² log x – x² + A
which is the required solution.

(ii) Given diff. eqn. be,
x \(\frac{d y}{d x}\) + 2y = x² log x
⇒ \(\frac{d y}{d x}+\left(\frac{2}{x}\right)\) y = x log x
which is L.D.E. and is of the form
\(\frac{d y}{d x}\) + Py = Q ;
where P = \(\frac{2}{x}\)
and Q = x log x
∴ I.F. = e^{∫ P dx}
= \(e^{\int \frac{2}{x} d x}\)
= e^{2 log x}
= e^{log x2}
= x²
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + c
⇒ y . x² = ∫ x log x . x² dx + c
⇒ x²y = log x . \(\frac{x^4}{4}-\int \frac{1}{x} \cdot \frac{x^4}{4}\) dx + c
⇒ x²y = \(\frac{x^4}{4} \log x-\frac{x^4}{16}\) + c
⇒ y = \(\frac{x^4}{16}\) [4 log |x| – 1] + \(\frac{c}{x^2}\)
which is the required solution.

Question 14.
(i) \(\frac{d y}{d x}\) + y cot x = 2 cos x
(ii) \(\frac{d y}{d x}\) – y tan x = 2 sin x
Solution:
(i) \(\frac{d y}{d x}\) + y cot x = 2 cos x ……………(1)
On comparing with \(\frac{d y}{d x}\) + Py = Q
where P = cot x ;
Q = 2 cos x ;
I.F. = e^{∫ cot x dx}
= e^{log |sin x|}
= sin x
Multiply eqn. (1) by sin x ; we get
sin x \(\frac{d y}{d x}\) + y cos x = 2 sin x cos x
⇒ \(\frac{d}{d x}\) (y sin x) = sin 2x
On integrating ; we have
y sin x = – \(\frac{\cos 2 x}{2}\) + C is the required solution.

(ii) Given, \(\frac{d y}{d x}\) – y tan x = 2 sin x
which is L.D.E. in y and is of the form
\(\frac{d y}{d x}\) + Py = Q
where P = – tan x
and Q = 2 sin x
∴ I.F. = e^{∫ P dx}
= ∫ Q . e^{∫ P dx} + C
⇒ y . cos x = ∫ 2 sin x cos x dx + C
⇒ y cos x = – 2 ∫ cos (- sin x) dxx + C
⇒ y cos x = – 2 \(\frac{\cos ^2 x}{2}\) + C
⇒ y cos x = – cos² x + C
⇒ y = – cos x + C sec x
which gives the required solution.

Question 15.
(i) (tan x) \(\frac{d y}{d x}\) + 2y = sec x
(ii) dy = x (x² – 2y) dx
Solution:
(i) Given eqn. be,
tan x \(\frac{d y}{d x}\) + 2y = sec x
⇒ \(\frac{d y}{d x}\) + (2 cot x) y = \(\frac{\sec x}{\tan x}\) ……………….(1)
Comparing eqn. (1) with \(\frac{d y}{d x}\) + Py = Q
P = 2 cot x ;
Q = \(\frac{\sec x}{\tan x}\)
= sec x × tan x
= cosec x
∴ I.F. = e^{∫ P dx}
= e^{∫ 2 cot x dx}
= \(e^{\int \frac{2 \cos x d x}{\sin x}}\)
= e^{2 log sin x}
= e^{log sin2 x}
= sin² x
Multiplying eqn. (1) by sin² x ; we get
sin² x \(\frac{d y}{d x}\) + 2 cot x sin² x y = cosec x . sin² x
∴ \(\frac{d}{d x}\) (y sin² x) = sin x ;
On integrating ; we have
y sin² x = – cos x + C is the required solution.

(ii) Given, dy = x (x² – 2y) dx
⇒ \(\frac{d y}{d x}\) + 2xy = x³
which is L.D.E. of the form \(\frac{d y}{d x}\) + Py = Q
where P = 2x
and Q = x³
∴ I.F. = e^{∫ P dx}
= e^{∫ 2x dx}
= e^x2
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
∴ y . e^x2 = ∫ x³ . e^x2 dx + C
put x² = t
⇒ 2x dx = dt
y e^x2 = ∫ e^t . t \(\frac{d t}{2}\) + C
ye^x2 = \(\frac{1}{2}\) [t e^t – e^t] + C
ye^x2 = \(\frac{1}{2}\) (x² – 1) e^x2 + C
y = \(\frac{1}{2}\) (x² – 1) + Ce^x2
which is the required solution.

Question 16.
(i) \(\frac{d y}{d x}\) – y = cot x
(ii) x log x \(\frac{d y}{d x}\) + y = 2 log x, x > 1
Solution:
(i) Given, \(\frac{d y}{d x}\) – y = cot x
which is linear in y and is of the form
\(\frac{d y}{d x}\) + Py = Q
where P = – 1 and Q = cos x
∴ I.F. = e^{∫ P dx}
= e^{∫ – dx}
= e^{– x}
and solution is given by
⇒ y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ ye^{– x} = 1 + C ……………….(1)
where I =∫ e^{– x} cos x dx
= \(\cos x \frac{e^{-x}}{-1}-\int(-\sin x) \frac{e^{-x}}{-1}\) dx
= \(-\cos x \cdot e^{-x}-\left[\sin x \frac{e^{-x}}{-1}+\int \cos x e^{-x} d x\right]\)
2I = (sin x – cos x) e^{– x}
∴ from (1) ; we have
y e^{– x} = \(\frac{e^{-x}}{2}\) (sin x – cos x) + c
⇒ y = \(\frac{1}{2}\) (sin x – cos x) + C e^x
which gives the required solution.

(ii) Given diff. eqn. be, x log x – \(\frac{d y}{d x}\) + y = 2 log x
⇒ \(\frac{d y}{d x}+\frac{1}{x \log x} y=\frac{2}{x}\)
which is linear differential eqn. and is of the form
\(\frac{d y}{d x}\) + Py = Q
where P = \(\frac{1}{x \log x}\)
and Q = \(\frac{2}{x}\)
Now
I.F. = e^{∫ P dx}
= \(e^{\int \frac{1 / x}{\log x} d x}\)
= e^{log (log x)}
= log x
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + c
⇒ y . log x = ∫ \(\frac{2}{x}\) log x dx + c
⇒ y log x = 2 \(\frac{(\log x)^2}{2}\) + c
[∵ ∫ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\) + c, n ≠ – 1]
⇒ y = log x + \(\frac{c}{llog x}\) be the required solution.

Question 17.
(i) y dx + (x – y²) dy = 0, y > 0 (NCERT)
(ii) (x + 3y²) dy = y dx (NCERT)
Solution:
(i) Given, y dx + (x – y²) dy = 0
⇒ \(\frac{d x}{d y}\) + \(\frac{x}{y}\) – y = 0
⇒ \(\frac{d x}{d y}+\frac{x}{y}\) = y
which is linear diff. eqn. in x and is of the form
\(\frac{d x}{d y}\) + Px = Q
where P = \(\frac{1}{y}\) and Q = y
∴ I.F. = e^{∫ P dy}
= \(e^{\int \frac{1}{y} d y}\)
= e^{log y} = y
and solution is given by
x . e^{∫ P dy} = ∫ Q . e^{∫ P dy} dy + C
⇒ xy = ∫ y . y dy + C
⇒ xy = \(\frac{y^3}{3}\) + C
which gives the required solution.

(ii) Given, (x + 3y²) dy = y dx
⇒ \(\frac{d x}{d y}=\frac{x}{y}\) + 3y
⇒ \(\frac{d x}{d y}-\frac{1}{y} x\) = 3y
which is linear diff. eqn. in x and is of the form
\(\frac{d x}{d y}\) + Px = Q
where P = – \(\frac{1}{y}\)
and Q = 3y
∴ I.F. = e^{∫ P dy}
= \(e^{\int-\frac{1}{y} d y}\)
= e^{– log y}
= e^{log y-1}
= \(\frac{1}{y}\)
and solution is given by
x . e^{∫ P dy} = ∫ Q . e^{∫ P dy} dy + C
⇒ x . \(\frac{1}{y}\) = ∫ 3y × \(\frac{1}{y}\) dy + C
⇒ \(\frac{x}{y}\) = 3y + C
⇒ x = 3y² + Cy, which gives the required solution.

Question 18.
(i) dx + x dy = e^{– y} log y dy, y > 0
(ii) (x + 2y³) dy = y dx, y > 0 (NCERT Exemplar)
Solution:
(i) Given, dx + x dy = e^{– y} log y dy
⇒ dx = (e^{– y} log y – x) dy
⇒ \(\frac{d x}{d y}\) + x = e^{– y} log y
which is L.D.E. in x and is of the form
\(\frac{d x}{d y}\) + Px = Q
where P = 1 ; Q = e^{– y} log y
∴ I.F. = e^{∫ P dy}
= e^{∫ dy}
= e^y
and solution is given by
x . e^{∫ P dy} = ∫ Q . e^{∫ P dy} dy + C
⇒ xe^y = ∫ e^{– y} log y . e^y dy + C
⇒ xe^y = ∫ log y . 1 dy + C
⇒ xe^y = y log y – y + C
⇒ xe^y = y (log y – 1) + C
which gives the required solution.

(ii) Given, (x + 2y³) dy = y dx
⇒ \(\frac{d x}{d y}=\frac{x+2 y^3}{y}\)
⇒ \(\frac{d x}{d y}-\frac{x}{y}\) = 2y²
which is linear diff. eqn. in x and is of the form
\(\frac{d x}{d y}\) + Px = Q ;
where P = – \(\frac{1}{y}\)
and Q = 2y²
∴ I.F. = e^{∫ P dy}
= \(e^{\int-\frac{1}{y} d y}\)
= e^{– log y}
= e^{log y-1}
= \(\frac{1}{y}\)
and solution is given by
x . e^{∫ P dy} = ∫ Q . e^{∫ P dy} dy + C
⇒ \(\frac{x}{y}\) = ∫ 2y² . \(\frac{1}{y}\) dy + C
⇒ \(\frac{x}{y}\) = y² + C
⇒ x = y² + Cy
which gives the required solution.

Question 19.
(tan^-1 y – x) dy = (1 + y²) dx
Solution:
Given diff. eqn. be
(1 + y²) dx = (tan^-1 y – x) dy

x e^{tan-1 y} = ∫ e^t . t dt + C
= (t – 1) e^t + C
⇒ x e^{tan-1 y} = (tan^-1 y – 1) e^{tan-1 y} + C
⇒ x = tan^-1 y – 1 + C e^{– tan-1 y}
which is the required solution.

Question 20.
Solve the differential equation (1 + x²) \(\frac{d y}{d x}\) + 2xy – 4x² = 0 subject to the initial condition y(0) = 0.
Solution:
Given eqn. is \(\frac{d y}{d x}\) + 2Y = 6e^x …………….(1)
Now eqn. (1) is LD.E of 1st order in y.
On comparing with \(\frac{d y}{d x}\) + Py = Q;
we have
P = 2 and Q = 6e^x
∴ I.F = e^{∫ P dx}
= e^{∫ 2 dx}
= e^2x
Multiply eqn. (1) by e^2x; we get
e^2x \(\frac{d y}{d x}\) + 2 e^2x y = 6 e^3x
⇒ \(\frac{d}{d x}\) (y e^2x) = 6 e^3x ;
On integrating we have
y e^2x = ∫ 6 e^3x dx + C
⇒ y e^2x = \(\frac{6 e^{3 x}}{3}\) + C
⇒ y = 2 e^x + C e^-2x is the required solution.

Question 21.
Solve the differential equation \(\frac{d y}{d x}\) + y cot x = 2x + x² cot x, given that y = 0 when x = \(\frac{\pi}{2}\).
Solution:
Given, \(\frac{d y}{d x}\) + y cot x = 2x + x² cot x ………………….(1)
On comparing with \(\frac{d y}{d x}\) + Py = Q ;
we have
P = cot x ; Q = 2x + x² cot x
∴ I.F. = e^{∫ P dx}
= \(e^{\int \frac{\cos x}{\sin x} d x}\)
Multiply eqn. (1) by sin x; we get
sin x \(\frac{d y}{d x}\) + y cos x = [2x sin x + x² cos x]
⇒ \(\frac{d}{d x}\) (y sin x) = 2x sin x + x² cos x;
on integrating ; we have
y sin x = ∫ 2x sin x dx + ∫ x² cos x dx + C
⇒ y sin x = ∫ 2x sin x dx + [x² sin x – ∫ 2x sin x dx] + C
⇒ y sin x = x² sin x + C ……………..(2)
given y = 0 when x = \(\frac{\pi}{2}\)
∴ from (2) ;
0 = (\(\frac{\pi}{2}\))² + C
⇒ C = – \(\frac{\pi^2}{4}\)
Thus eqn. (2) givcs;
y sin x = x² sin x – \(\frac{\pi^2}{4}\)
⇒ y = x² – \(\frac{\pi}{4}\) cosec x.

Question 22.
If \(\frac{d y}{d x}\) + 2xy = x, then prove taht 2y = 1 + e^{– x2}, given that y = 1 when x = 0.
Solution:
Given, \(\frac{d y}{d x}\) + 2xy = x,
which is L.D.E. in y and is of the form
\(\frac{d y}{d x}\) + Py = Q
∴ I.F. = e^{∫ P dx}
= e^{∫ 2x dx}
= e^x2
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ y . e^x2 = ∫ x . e^x2 dx + C
put x² = t
⇒ 2x dx = dt
y . e^x2 = \(\frac{1}{2}\) ∫ e^t dt
⇒ y . e^x2 = \(\frac{1}{2}\) e^x2 + C
⇒ y = \(\frac{1}{2}\) + Ce^-x2 ………………(1)
Given y = 1, when x = 0
∴ from (1) ;
1 = \(\frac{1}{2}\) + Ce⁰
⇒ C = \(\frac{1}{2}\)
∴ from (1) ; we have
2y = 1 + e^-x2 be the required solution.

Question 23.
Solve (x + 2y²) \(\frac{d y}{d x}\) = y, given that x = 2 when y = 1.
Solution:
Given diff. eqn. be,
(x + 2y²) \(\frac{d y}{d x}\) = y
⇒ \(\frac{d x}{d y}=\frac{x+2 y^2}{y}=\frac{x}{y}\) + 2y
⇒ \(\frac{d x}{d y}-\frac{x}{y}\) = 2y
which is linear differential equation of the form \(\frac{d x}{d y}\) + Px = Q
where P = – \(\frac{1}{y}\)
and Q = 2y
Now I.F. = e^{∫ P dy}
= \(e^{\int-\frac{1}{y} d y}\)
= e^{– log |y|}
= \(\frac{1}{y}\)
x.e^{∫ P dy} = ∫ Q . e^{∫ P dy} dy + C
⇒ \(\frac{x}{y}\) = ∫ 2y . \(\frac{1}{y}\) dy + c
⇒ \(\frac{x}{y}\) = 2y + c
⇒ x = 2y² + cy ……………..(1)
Given where x = 2, y = 1
∴ From (1) ;
2 = 2 + c
⇒ c = 0
∴ From eqn. (1) ; we have
x = 2y² be the required solution.

Question 24.
Solve the differential equation :
\(\frac{d y}{d x}\) + y sec² x = tan x sec² x, y(0) = 1.
Solution:
Given, \(\frac{d y}{d x}\) + y sec² x = tan x sec² x
which is L.D.E. in y and is of the form
\(\frac{d y}{d x}\) + Py = Q ;
where P = sec² x ;
Q = tan x sec² x
∴ I.F. = e^{∫ P dx}
= e^{∫ sec2 x dx}
= e^{tan x}
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ y . e^{tan x} = ∫ tan x . sec² x . e^{tan x} dx + C
put tan x = t
⇒ sec² x dx = dt
y e^{tan x} = ∫ t e^t dt + C
⇒ y e^{tan x} = [t e^t – e^t] + C
⇒ y . e^{tan x} = (tan x – 1) e^{tan x} + C
⇒ y = (tan x – 1) + C e^{– tan x} ……………….(1)
which gives the required general solution.
Given y(0) = 1 i.e. when x = 0 ; y = 1
∴ from (1) ;
1 = – 1 + C
⇒ C = 2
Thus eqn. (1) gives ;
y = (tan x – 1) + 2e^{– tan x}
which gives the required solution..

Question 25.
Find the particular solutions of the following differential equations:
(i) y’ – y = e^x, given that y = 1 when x = 0
(ii) y’ + y = e^x, given that y = \(\frac{1}{2}\) when x = 0
(iii) xy’ + y = x log x, given that y = \(\frac{1}{4}\) when x = 1
(iv) xy’ – y = log x, given that y = 0 when x = 1.
(v) (1 + x²) \(\frac{d y}{d x}\) + 2xy = \(\frac{1}{1+x^2}\), given that y = 0 when x = 1.
(vi) \(\frac{d y}{d x}\) + 2y tan x = sin x, given that y = 0 when x = \(\frac{\pi}{3}\).
(vii) y’ + 2y = e^-2x sin x, given that y = 0 when x = 0.
(viii) xy’ + y = x cos x + sin x, given that y = 1 when x = \(\frac{\pi}{2}\).
Solution:
(i) Given y’ – y = e^x,
which is L.D.E. in y and is of the form
\(\frac{d y}{d x}\) + Py = Q ;
where P = – 1 and Q = e^x
∴ I.F. = e^{∫ P dx}
= e^{∫ – dx}
= e^{– x}
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ ye^{– x} = ∫ e^x . e^{– x} dx + C
⇒ ye^{– x} = x + C ………………(1)
Given that y = 1 when x = 0
∴ from (1) ;
1 . e^-0 = 0 + C
⇒ C = 1
Thus eqn. (1) gives ;
y e^{– x} = x + 1
⇒ y = (x + 1) e^x
which gives the required solution.

(ii) Given diff. eqn. be,
y’ + y = e^x
which is L.D.E. and of the form
\(\frac{d y}{d x}\) + Py = Q
where p = 1 and Q = e^x
∴ I.F. = e^{∫ P dx}
= e^{∫ dx} dx
= e^x
and its solution is given by
y . e^{∫ P dx} =∫ Q . e^{∫ P dx} dx + c
⇒ y . e^x = ∫ e^x . e^x dx + c
= ∫ e^2x dx + c
⇒ ye^x = \(\frac{e^{2 x}}{2}\) + c
y = \(\frac{e^x}{2}\) + ce^-x ………………(1)
Given y (0) = \(\frac{1}{2}\)
i.e. when x = 0 ; y = \(\frac{1}{2}\)
∴ From eqn. (1) ;
we have \(\frac{1}{2}\) = \(\frac{1}{2}\) + c
⇒ c = 0
∴ From eqn. (1) ;
y = \(\frac{e^x}{2}\),
be the required solution of initial value problem.

(iii) Given, xy’ + y = x log x
⇒ \(\frac{d y}{d x}\) + \(\frac{1}{x}\) y = log x
which is L.D.E. in y of first order
and is of the form,
\(\frac{d y}{d x}\) + Py = Q
where P = \(\frac{1}{x}\) and Q = log x
∴ I.F. = e^{∫ P dx}
= \(e^{\int \frac{1}{x} d x}\)
= e^{log x} = x
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ y . x = ∫ log x . x dx + C
⇒ xy = log x . \(\frac{x^2}{2}\) – \(\int \frac{1}{x} \cdot \frac{x^2}{2}\) dx + C
⇒ xy = \(\frac{x^2}{2}\) log x – \(\frac{x^2}{4}\) + C
⇒ y = \(\frac{x}{2}\) log x – \(\frac{x}{4}\) + \(\frac{C}{x}\) …………..(1)
Given that y = \(\frac{1}{4}\) when x = 1
∴ from (1) ;
\(\frac{1}{4}\) = \(\frac{1}{2}\) log 1 – \(\frac{1}{4}\) + C
⇒ C = \(\frac{1}{2}\)
∴ from (1) ; we have
y = \(\frac{x}{2}\) log x – \(\frac{x}{4}\) + \(\frac{1}{2 x}\),
which gives the required solution.

(iv) Given diff. eqn. be,
x \(\frac{d y}{d x}\) – y = log x
⇒ \(\frac{d y}{d x}-\frac{y}{x}=\frac{\log x}{x}\)
which is L.D.E. of the form
\(\frac{d y}{d x}\) + Py = Q
where P = – \(\frac{1}{x}\)
and Q = \(\frac{log x}{x}\)
I.F. = e^{∫ P dx}
= \(e^{\int-\frac{1}{x} d x}\)
= e^{– log x}
= e^{log x-1}
= \(\frac{1}{x}\)
and solution is given by
ye^{∫ P dx} = ∫ Q . e^{∫ P dx} + c
⇒ y . \(\frac{1}{x}\) = \(\int \frac{\log x}{x} \cdot \frac{1}{x} d x\) + c
put log x = t
⇒ x = e^t
⇒ dx = e^t dt
= \(\int \frac{t}{e^{2 t}}\) e^t dt + c
= ∫ t . e^-t dt + c
⇒ \(\frac{y}{x}\) = [- t e^-t – e^-t] + c
⇒ y = – x (log x + 1) + cx ………………..(1)
Given y(1) = 0 i.e. when x = 1 ; y = 0
∴ from (1) ; we have
0 = – (0 + 1) + c
c = 1
∴ From eqn. (2) ; we have
y = – (log x + 1) + x be the reqd. solution.

(v) Given diff. eqn. can be written as
\(\frac{d y}{d x}+\frac{2 x}{1+x^2} y=\frac{1}{\left(1+x^2\right)^2}\)
which is L.D.E. in y of first order and comparing with
\(\frac{d y}{d x}\) + Py = Q
where P = \(\frac{2 x}{1+x^2}\) ;
Q = \(\frac{1}{\left(1+x^2\right)^2}\)
∴ I. F. = e^{∫ P dx}
= \(e^{\int \frac{2 x}{1+x^2} d x}\)
= e^{log (1 + x2)}
= (1 + x²)
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} + C
y (1 + x²) = tan^-1 x + C …………………(1)
Given y (0) = 0 i.e. when x = O, y = O
∴ from (2) ;
0 = 0 + C
⇒ C = 0
∴ eqn. (2) becomes ;
y(1 + x²) = tan^-1 x is the required solution.

(vi) Given \(\frac{d y}{d x}\) + 2y tan x = sin x
which is linear diff. eqn of the form
\(\frac{d y}{d x}\) + Py = Q;
where P = 2 tan x and Q = sin x
∴ I.F. = e^{∫ P dx}
= e^{∫ tan x dx}
= e^{– 2 log |cos x|}
and solution is given by
ye^{∫ P dx} = ∫ Q . e^{∫ P dx} dx +c
⇒ y . sec² x = ∫ sin x . sec² x dx + c
⇒ y sec² x = ∫ tan x sec x dx + c
⇒ y sec² x = sec x + c ……………………..(1)
Given y = 0 when x = \(\frac{\pi}{3}\)
∴ eqn (1) gives;
0 = 2 + c
⇒ c = -2
From eqn (1); we have
y sec² x = sec x – 2
⇒ y = \(\frac{1}{\sec x}-\frac{2}{\sec ^2 x}\)
y = cos x – 2cos² x
which is the requried solution.

(vii) Given diff. eqn be,
\(\frac{d y}{d x}\) + 2y = e^{– 2x} sin x
which is L.D.E. of the form \(\frac{d y}{d x}\) + Py = Q
where P = 2
and Q = e^{– 2x} sin x
Now. I.F. = e^{∫ P dx}
= e^{∫ 2 dx}
= e^2x
and solution is given by
y . e^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + c
⇒ y . e^2x = ∫ e^-2x sin x . e^2x dx + c
⇒ ye^2x = – cos x + c ……………..(1)
Given y(0) = 0 i.e. when x = 0 ; y = 0
∴ From (1);
0 = – 1 + c
⇒ c = 1
∴ From eqn (1) ; we have
ye^2x = 1 – cos x, which is the required solution of given initial value problem.

(viii) Given x \(\frac{d y}{d x}\) + y = x cos x + sin x
⇒ \(\frac{d y}{d x}+\frac{y}{x}=\cos x+\frac{\sin x}{x}\)
which is L.D.E. of the form \(\frac{d y}{d x}\) + Py = Q
where P = \(\frac{1}{x}\)
and Q = cos x + \(\frac{sin x}{x}\)
∴ I.F. = e^{∫ P dx}
= \(e^{\int \frac{1}{x} d x}\)
= e^{log x}
= x
and solution is given by
ye^{∫ P dx} = ∫ Q . e^{∫ P dx} dx + C
⇒ yx = ∫ [cos x + \(\frac{sin x}{x}\)] x dx + c
⇒ xy = ∫ x cos x dx + ∫ sin x dx + c
= x sin x – ∫ sin x dx – cos x + c
= x sin x + cos x – cos x + c
⇒ xy = x sin x + c ……………..(1)
Given y(\(\frac{\pi}{2}\)) = 1
i.e. when x = \(\frac{\pi}{2}\) ; y = 1
∴ From eqn (1) ;
\(\frac{\pi}{2}\) × 1 = \(\frac{\pi}{2}\) × 1 + c
⇒ c = 0
∴ From eqn (1) ; we have
xy = x sin x
⇒ y = sin x be the required solution.

Accessing Class 12 ISC Maths Solutions Chapter 9 Differential Equations Ex 9.6 can be a valuable tool for students seeking extra practice.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.6

Very Short answer type questions (1 to 2) :

Question 1.
Which of the following differential equations are homogeneous ?
(i) (x – y) \(\frac{d y}{d x}\) = x + 2y
(ii) y – x \(\frac{d y}{d x}\) = x + y \(\frac{d y}{d x}\)
Solution:
(i) Given (x – y) \(\frac{d y}{d x}\) = x + 2y
⇒ \(\frac{d y}{d x}=\frac{x+2 y}{x-y}\),
which is homogenenous diff. eqn.
Since \(\frac{d y}{d x}=\frac{x\left(1+\frac{2 y}{x}\right)}{x\left(1-\frac{y}{x}\right)}=\phi\left(\frac{y}{x}\right)\)
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
Thus from given eqn ; we have

(ii) Given differential eqn. be \(\frac{d y}{d x}=\frac{y-x}{y+x}\)
which is homogenenous diff. equation.
Since \(\frac{d y}{d x}=\frac{x\left(\frac{y}{x}-1\right)}{x\left(\frac{y}{x}-1\right)}=\phi\left(\frac{y}{x}\right)\)
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ From eqn. (1) ; we have

⇒ \(\frac{1}{2}\) log (v² + 1) + tan^-1 v + log x = \(\frac{1}{2}\) log c
⇒ log (v² + 1) + tan^-1 v + log x = \(\frac{1}{2}\) log c
⇒ log (\(\frac{y^2}{x^2}\) + 1) x² + 2 tan^-1 (\(\frac{y}{x}\)) = A
⇒ log (x² + y²) + 2 tan^-1 (\(\frac{y}{x}\)) = A
which is the required solution.

Question 1 (old).
When a differential equation is called homogeneous ?
Solution:
A differential eqn. \(\frac{d y}{d x}\) = f(x, y) is said to be homogeneous diff. eqn. iff(x, y) be a homogenenous function of degree 0.
i.e. a homogenenous diff. eqn. is of the form,
\(\frac{d y}{d x}\) = f (\(\frac{y}{x}\))
or \(\frac{d x}{d y}\) = g (\(\frac{x}{y}\))

Question 2.
(x² + 3xy – 4y²) dx – x (x² + 2y) dy = 0
Solution:
Given diff. eqn. can be written as,
\(\frac{d y}{d x}=\frac{x^2+3 x y-4 y^2}{x\left(x^2+2 y\right)}\)
= \(\frac{x^2\left[1+\frac{3 y}{x}-4\left(\frac{y}{x}\right)^2\right]}{x^2\left(x+\frac{2 y}{x}\right)}\)
i.e. \(\frac{d y}{d x}\) is not a function of \(\frac{y}{x}\).
Thus given diff. eeqn. is not homogenenous.

Question 3.
(i) x \(\frac{d y}{d x}\) + y = x tan^-1 \(\frac{y}{x}\)
(ii) x cos (\(\frac{y}{x}\)) \(\frac{d y}{d x}\) = y cos (\(\frac{y}{x}\)) – 3x²
Solution:
(i) Given diff. eqn. be,
x \(\frac{d y}{d x}\) + y = x tan^-1 \(\frac{y}{x}\)
⇒ \(\frac{d y}{d x}\) = – \(\frac{y}{x}\) + tan^-1 \(\frac{y}{x}\)
Thus \(\frac{d y}{d x}\) = f(\(\frac{y}{x}\))
∴ given diff. eqn. be homogenenous diff. eqn.

(ii) Given diff. eqn. be,
x cos (\(\frac{y}{x}\)) \(\frac{d y}{d x}\) = y cos (\(\frac{y}{x}\)) – 3x²
∴ \(\frac{d y}{d x}=\frac{y \cos \left(\frac{y}{x}\right)-3 x^2}{x \cos \left(\frac{y}{x}\right)}\)
= \(\frac{\frac{y}{x} \cos \left(\frac{y}{x}\right)-3 x}{\cos \left(\frac{y}{x}\right)}\)
which is not of the form, \(\frac{d y}{d x}\) = f(\(\frac{y}{x}\))
Hence given diff. eqn. is not homogenenous.

Question 3 (old).
(ii) (x – y) dy – (x + y) dx = 0 (NCERT)
Solution:
Given \(\frac{d y}{d x}=\frac{x+y}{x-y}\),
which is homogeneous diff. eqn.
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
Thus given eqn. becomes

Question 4.
(i) x \(\frac{d y}{d x}\) – y = \(\sqrt{x^2+y^2}\)
(ii) (x² + y² sin \(\frac{x}{y}\)) \(\frac{d y}{d x}\) = 2xy e^x/y
Solution:
(i) Given, diff. eqn. be
x \(\frac{d y}{d x}\) – y = \(\sqrt{x^2+y^2}\)
⇒ \(\frac{d y}{d x}=\frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^2}\)
= f (\(\frac{y}{x}\))
Hence given diff. eqn. is homogeneous diff. eqn.

(ii) Given diff. eqn. be

Question 5.
(i) 2y e^{\(\frac{x}{y}\)} dx + (y – 2xe^{\(\frac{x}{y}\)}) dy = 0
(ii) y dx + x log (\(\frac{y}{x}\)) dy – 2x dy = 0
Solution:
(i) Given, diff. eqn. be
2y e^{\(\frac{x}{y}\)} dx + (y – 2xe^{\(\frac{x}{y}\)}) dy = 0
⇒ \(\frac{d x}{d y}=-\frac{\left(y-2 x e^{x / y}\right)}{2 y e^{x / y}}\)
= \(\frac{\frac{2 x}{y} e^{x / y}-1}{2 e^{x / y}}\)
which is of the form \(\frac{d x}{d y}\) = f (\(\frac{x}{y}\))
Hence given diff. eqn. be homogenenous.

(ii) Given differential eqn. can be written as
y dx = [2x – x log (\(\frac{y}{x}\))] dy
\(\frac{d x}{d y}=\frac{2 x}{y}+\frac{x}{y} \log \frac{x}{y}\)
= f (\(\frac{x}{y}\))
Thus, given differential eqn. be homogenenous.

Question 6.
(i) \(\frac{d y}{d x}=\frac{x+y}{x}\) (NCERT)
(ii) (x – y) dy = (x + y) dx
Solution:
(i) Given diff. eqn. be,
\(\frac{d y}{d x}=\frac{x+y}{x}\)
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ v + x \(\frac{d v}{d x}\) = \(\frac{x+v x}{x}\) = 1 + v
x \(\frac{d v}{d x}\) = 1
⇒ dv = \(\frac{d x}{x}\)
On integrating ; we have
v = log x + C
y = x log |x| + Cx
which is the required solution.

(ii) Given \(\frac{d y}{d x}=\frac{x+y}{x-y}\),
which is homogenenous diff. eqn.
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
Thus, given eqn becomes ;

Question 7.
(i) (x – y) y’ = x + 3y
(ii) (x + 2y) dx – (2x – y) dy = 0
Solution:
(i) Given, (x – y) y’ = x + 3y
⇒ y’ = \(\frac{x+3 y}{x-y}\) …………….(1)
Also, y’ = \(\frac{d y}{d x}\)
= \(\frac{1+3 \frac{y}{x}}{1-\frac{y}{x}}\)
= f(\(\frac{y}{x}\) )
Thus eqn. (1) be homogenenous diff. eqn.
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ from (1) ; we have

(ii) Given (x + 2y) dx – (2x – y) dy = 0
⇒ \(\frac{d y}{d x}=\frac{x+2 y}{2 x-y}\),
which is homogenenous differential equation.
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)

Question 7 (old).
Prove that x² – y² = c (x² + y²)² is the general solution of the differential equation (x³ – 3xy²) dx = (y³ – 3x²y) dy where c is a parameter.
Solution:
Given differential eqn. can be written as
\(\frac{d y}{d x}=\frac{x^3-3 x y^2}{y^3-3 x^2 y}\) ………………….(1)
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ from (1) ;

Question 8.
(i) (y² – 2xy) dx = (x² – 2xy) dy
(ii) 2xy dx + (x² + 2y²) dy = 0
Solution:
(i) Given, (y² – 2xy) dx = (x² – 2xy) dy
⇒ \(\frac{d y}{d x}=\frac{y^2-2 x y}{x^2-2 x y}\) ……………..(1)
which is homogenenous differential equation
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ From eqn. (1) ; we have

⇒ – log |v| – log |v – 1| = 3 log |x| + log c
log [v (v – 1) x³] = – log c
= log \(\frac{1}{c}\)
= log A
⇒ \(\frac{y}{x}\left(\frac{y}{x}-1\right)\) x³ = A
⇒ \(\frac{y(y-x)}{x^2}\) × x³ = A
⇒ xy² – x²y = A be the required solution.

(ii) Given 2xy dx + (x² + 2y²) dy = 0
⇒ \(\frac{d y}{d x}=-\frac{2 x y}{x^2+2 y^2}\) ……………….(1)
which is homogenenous differential equation
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ From eqn. (1) ; we have

Question 8 (old).
(ii) x \(\frac{d y}{d x}\) = y (log y – log x + 1)
Solution:
Given \(\frac{d y}{d x}\) = \(\frac{y}{x}\) (log y – log x + 1)
⇒ \(\frac{d y}{d x}=\frac{y}{x}\left\{\log \frac{y}{x}+1\right\}\) …………………….(1)
[∵ log a – log b = log \(\frac{a}{b}\)]
putting y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
Thus from eqn. (1) ; we have
v + x \(\frac{d v}{d x}\) = v {log v + 1}
⇒ x \(\frac{d v}{d x}\) = v log v
\(\frac{d v}{v \log v}=\frac{d x}{x}\)
integrating both sides, we have
\(\int \frac{d v}{v \log v}=\int \frac{d x}{x}\)
⇒ \(\int \frac{\frac{1}{v} d v}{\log v}=\int \frac{d v}{x}\)
⇒ log (log v) – log x = log c
⇒ log [log v] = log cx
⇒ log v = cx
⇒ log (y/x) = cx be the required solution.

Question 9.
(i) (x² y²) dx + 2xy dy = 0 (NCERT)
(ii) x² \(\frac{d y}{d x}\) = x² + xy + y² (NCERT Exemplar)
Solution:
(i) Given, \(\frac{d y}{d x}=\frac{y^2-x^2}{2 x y}\) ………….(1)

(ii) Given x² \(\frac{d y}{d x}\) = x² + xy + y²
⇒ \(\frac{d y}{d x}=\frac{x^2+x y+y^2}{x^2}\) ………………..(1)
which is homogenenous differential equation
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ From eqn. (1) ; we have
v + x \(\frac{d v}{d x}\) = \(\frac{x^2+v x^2+v^2 x^2}{x^2}\)
⇒ x \(\frac{d v}{d x}\) = 1 + v + v² – v
= 1 + v²
⇒ \(\frac{d v}{1+v^2}=\frac{d x}{x}\) ;
on integrating
⇒ \(\int \frac{d v}{1+v^2}=\int \frac{d x}{x}\)
⇒ tan^-1 v = log |x| + c
⇒ tan^-1 \(\frac{y}{x}\) = log |x| + c be the required solution.

Question 10.
(i) x \(\frac{d y}{d x}\) + \(\frac{y^2}{x}\) = y
(ii) 3x² \(\frac{d y}{d x}\) – 3xy = y²
Solution:
(i) Given, x \(\frac{d y}{d x}\) + \(\frac{y^2}{x}\) = y
⇒ x \(\frac{d y}{d x}\) = y – \(\frac{y^2}{x}\)
= \(\frac{x y-y^2}{x}\)
⇒ \(\frac{d y}{d x}=\frac{x y-y^2}{x}\) ………………(1)
Clearly \(\frac{d y}{d x}=\frac{y}{x}-\left(\frac{y}{x}\right)^2=f\left(\frac{y}{x}\right)\)
Thus, eqn. (1) be homogenenous differential equation.
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ from (1) ;
v + x \(\frac{d v}{d x}\) = \(\frac{v x^2-v^2 x^2}{x^2}\)
⇒ x \(\frac{d v}{d x}\) = v – v² – v = – v²
⇒ \(\frac{d v}{v^2}=-\frac{d x}{x}\)
on integrating ; we have
⇒ \(\int \frac{d v}{v^2}=-\int \frac{d x}{x}+\mathrm{C}\)
⇒ – \(\frac{1}{v}\) = – log |x| + C
⇒ – \(\frac{x}{y}\) = – log |x| + C
⇒ log |x| = C + \(\frac{x}{y}\)
⇒ |x| = e^x/y e^c
⇒ x = ± e^c e^x/y
= A e^x/y which is the required solution.

(ii) Given diff. eqn. be written as ;
\(\frac{d y}{d x}=\frac{3 x y+y^2}{3 x^2}\) …………..(1)
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ from (1) ;

Question 11.
(i) \(\frac{d y}{d x}=\frac{y}{x}+\frac{\sqrt{x^2-y^2}}{x}\), x > 0
(ii) \(x \frac{d y}{d x}-y=2 \sqrt{y^2-x^2}\), x > 0
Solution:
(i) Given,
\(\frac{d y}{d x}=\frac{y}{x}+\frac{\sqrt{x^2-y^2}}{x}\), x > 0 ……………….(1)
Also, \(\frac{d y}{d x}=\frac{y}{x}+\sqrt{1-\left(\frac{y}{x}\right)^2}=\phi\left(\frac{y}{x}\right)\)
Thus eqn. (1) be homogenenous diff. eqn.
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ from (1) ;
v + x \(\frac{d v}{d x}\) = v + \(\sqrt{1-v^2}\)
⇒ \(x \frac{d v}{d x}=\sqrt{1-v^2}\)
⇒ \(\frac{d v}{\sqrt{1-v^2}}=\frac{d x}{x}\)
On integrating ; we have
\(\int \frac{d v}{\sqrt{1-v^2}}=\int \frac{d x}{x}+\mathrm{A}\)
⇒ sin^-1 v = log |x| + A
⇒ sin^-1 (\(\frac{y}{x}\)) = log |x| + A
which is the required solution.

(ii) Given,
\(x \frac{d y}{d x}-y=2 \sqrt{y^2-x^2}\), x > 0

Question 12.
\(\left(\frac{y}{x} \cos \frac{y}{x}\right)\)dx – \(\left(\frac{x}{y} \sin \frac{y}{x}+\cos \frac{y}{x}\right)\) dy = 0
Solution:
Given,
\(\left(\frac{y}{x} \cos \frac{y}{x}\right)\)dx – \(\left(\frac{x}{y} \sin \frac{y}{x}+\cos \frac{y}{x}\right)\) dy = 0

Question 13.
(i) (1 + e^y/x) dy + e^y/x (1 – \(\frac{y}{x}\)) dx = 0
(ii) 2ye^x/y dx – (y – 2xe^x/y) dy = 0
Solution:
(i) Given diff. eqn. can be written as ;
\(\frac{d y}{d x}=\frac{-e^{y / x}\left(1-\frac{y}{x}\right)}{1+e^{y / x}}\) ………………(1)
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ from (1) ; we have
v + x \(\frac{d v}{d x}\) = \(\frac{-e^v(1-v)}{1+e^v}\)
⇒ x \(\frac{d v}{d x}\) = \(\frac{-e^v(1-v)}{1+e^v}\) – v
= \(\frac{-e^v+v e^v-v-v e^v}{1+e^v}\)
⇒ \(\frac{\left(1+e^v\right) d v}{v+e^v}=-\frac{d x}{x}\)
On integrating both sides ; we have
\(\int \frac{\left(1+e^v\right) d v}{v+e^v}=-\int \frac{d x}{x}\) + log C
⇒ log |v + e^v| = – log |x| + log C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(d)}\) = log |f(x)| + C]
⇒ log |(v + e^v)x| = log C
⇒ log |y + x e^y/x| = log C
⇒ y + x e^y/x = A [∵ ± C = A]
which is the required solution.

(ii) Given diff. eqn. can be written as
\(\frac{d x}{d y}=\frac{\left(2 x e^{x / y}-y\right)}{2 y e^{x / y}}\) …………….(1)
Here f (x, y) = \(\frac{2 x e^{x / y}-y}{2 y e^{x / y}}\)
∴ f (kx, ky) = \(\frac{2 k x e^{k x / k y}-k y}{2 k y e^{k x / k y}}\) = k⁰ f (x, y)
∴ f (x, y) is a homogenenous function of degree 0.
put \(\frac{x}{y}\) = v
⇒ x = yv
Diff. both sides w.r.t. y, we have
⇒ \(\frac{d x}{d y}=v+y \frac{d v}{d y}\) ……………….(2)
Putting eqn. (2) in eqn. (1) ; we get

Question 14.
Solve the differential equation (x + y) dy + (x – y) dx = 0, given that y = 1 when x = 1. (NCERT)
Solution:
Given \(\frac{d y}{d x}=\frac{y-x}{x+y}\) ……………..(1)
Here f (x, y) = \(\frac{y-x}{x+y}\)
∴ f(kx, ky) = \(\frac{k y-k x}{k x+k y}\)
= \(\frac{k(y-x)}{k(x+y)}\)
= k⁰ f (x, y)
Thus f (x, y) is homogenenous function of degree 0.
putting y = vx so that
\(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\) ……………..(2)
using eqn. (2) in eqn. (1) ; we get

Question 14 (old).
Find particular solutions of the following differential equations :
(i) 2xy + y² – 2x² = 0, given that y(1) = 2.
(ii) \(\frac{d y}{d x}-\frac{y}{x}\) + cosec \(\left(\frac{y}{x}\right)\) = 0 given that y = 0 when x = 1.
(iii) xe^y/x – y + x \(\frac{d y}{d x}\) = 0, given that y(e) = 0.
Solution:
(i) Given 2xy + y² – 2x² = 0
⇒ \(\frac{d y}{d x}=\frac{2 x y+y^2}{2 x^2}\) ……………(1)
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\) ……………(2)
Using eqn. (2) in eqn. (1) ; we have
v + x \(\frac{d v}{d x}\) = \(\frac{2 v x^2+v^2 x^2}{2 x^2}\)
⇒ \(\frac{x d v}{d x}=\frac{2 v+v^2}{2}-v=\frac{v^2}{2}\)
On variable separation, we have
\(\frac{2}{v^2} d v=\frac{d x}{x}\) ;
On integrating
– \(\frac{2}{v}\) = log |x| + C
⇒ – \(\frac{2x}{y}\) = log |x| + C …………….(1)
given y(1) = 2 i.e. when x = 1, y = 2
∴ from (1) ;
– 1 = C
∴ eqn. (1) becomes
– \(\frac{2x}{y}\) = log |x| – 1
⇒ \(\frac{2x}{y}\) = 1 – log |x|
⇒ y = \(\frac{2 x}{1-\log |x|}\) ; x ≠ 0, 1
which is the required solution.

(ii) Given \(\frac{d y}{d x}-\frac{y}{x}\) + cosec \(\left(\frac{y}{x}\right)\) = 0
⇒ \(\frac{d y}{d x}\) = \(\frac{y}{x}\) – cosec \(\frac{y}{x}\) ……………(1)
which is homogenenous diff. eqn.
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ from (1) ; we have
v + x \(\frac{d v}{d x}\) = v – cosec v
⇒ x \(\frac{d v}{d x}\) = – cosec v
⇒ \(\frac{d v}{\ {cosec} v}=-\frac{d x}{x}\) ;
on integrating ; we have
⇒ ∫ sin v dv = – ∫ \(\frac{d v}{x}\)
⇒ – cos v = – log |x| + c
⇒ – cos \(\frac{y}{x}\) = – log |x| + c
Since y (1) = 0
i.e. when x = 1 ; y = 0
∴ from (2) ; we have
– 1 = 0 + c
⇒ c = – 1
∴ From (2) ; we have
1 – cos \(\frac{y}{x}\) = – log |x|
⇒ cos (\(\frac{y}{x}\)) – 1 = log |x| which is the required solution.

(iii) Given xe^y/x – y + x \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=\frac{y-x e^{y / x}}{x}\)
= \(\frac{y}{x}\) – e^{\(\frac{y}{x}\)} ………………(1)
which is homogeneous diff. eqn.
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ From (2) ;
– 1 + log |e| = c
from (1) ; we have
v + x \(\frac{d v}{d x}\) = v – e^v
⇒ x \(\frac{d v}{d x}\) = – e^v
⇒ e^{– v} dv = – \(\frac{d x}{x}\) ;
On integrating ; we have
∫ e^{– v} dv = – ∫\(\frac{d x}{x}\)
⇒ – e^{– v} + log |x| = c
⇒ – e^{– y/x} + log |x| = c ……………..(2)
Since y (e) = 0
i.e. When x = e ; y = 0
∴ From (2) ;
– 1 + log |e| = c
⇒ c = 0
∴ From (2) ;
e^{– y/x} = log |x|
⇒ – \(\frac{y}{x}\) = log (log |x|)
⇒ y = -x log (log |x|)
which is the required solution.

Question 15.
Solve the differential equation (x² – y²) dx + 2xy dy = 0, given that y = 1 when x = 1.
Solution:
Given diff. eqn. can be written as

⇒ log (1 + y²) + log x = log C
⇒ log x (1 + y²) = log C
⇒ x \(\left(1+\frac{y^2}{x^2}\right)\) = C
which is the required general solution.
Given y = 1 when x = 1
∴ from (2) ;
1 + 1 = C
⇒ C = 2
Thus eqn. (2) becomes ;
x² + y² = 2x, which is the required solution.

Question 16.
Find particular solutions of the following differential equations:
(i) (x² + y²) dx = 2xy dy, given that y = 0 when x = 1.
(ii) \(\frac{d y}{d x}=\frac{x y}{x^2+y^2}\) given that y = 1 when x = 0.
(iii) 2xy + y² – 2x² \(\frac{d y}{d x}\) = 0, given that y (1) = 2.
(iv) (x² + 3xy + y²) dx – x² dy = 0 given that y = 0 when x = 1.
(v) \(\frac{d y}{d x}-\frac{y}{x}\) + cosec (\(\frac{y}{x}\)) = 0, given that y = 0 when x = 1.
(vi) x \(\frac{d y}{d x}\) = y – x tan \(\left(\frac{y}{x}\right)\), given that y = \(\frac{\pi}{4}\) at x = 1.
(vii) x e^y/x – y + x \(\frac{d y}{d x}\) = 0, given that y(e) = 0
Solution:
(i) Given diff. eqn. can be written as ;
\(\frac{d y}{d x}=\frac{x^2+y^2}{2 x y}\) ………………….(1)
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
Thus eqn. (1) becomes ;

⇒ log |1 – v²| = log |x| – log C
⇒ log |(1 – v²)x| = log C
⇒ (1 – \(\frac{y^2}{x^2}\)) x = A
⇒ x² – y² = Ax ……………….(1)
Given y = 0 when x = 1
∴ from (1) ; A = 1
Thus eqn. (1) becomes ;
x² – y² = x
which is the required particular solution.

(ii) Given \(\frac{d y}{d x}=\frac{x y}{x^2+y^2}\) ………………(1)
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
in eqn. (1) ; we have

(iii) Given 2xy + y² – 2x² \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=\frac{2 x y+y^2}{2 x^2}\) …………….(1)
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
Using eqn. (2) in eqn. (1) ; we have

(iv) Given diff. eqn. can be written as ;
\(\frac{d y}{d x}=\frac{x^2+3 x y+y^2}{x^2}\) ……………..(1)
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
in eqn. (1) ; we have

(v) Given \(\frac{d y}{d x}-\frac{y}{x}\) + cosec (\(\frac{y}{x}\)) = 0
⇒ \(\frac{d y}{d x}\) = \(\frac{y}{x}\) – cosec \(\frac{y}{x}\) ……………..(1)
which is homogeneous diff. eqn.
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ from (1) ; we have
v + x \(\frac{d v}{d x}\) = v – cosec v
⇒ x \(\frac{d v}{d x}\) = – cosec v
⇒ \(\frac{d v}{\ {cosec} v}=-\frac{d x}{x}\) ;
on integrating ; we have
⇒ ∫ sin v dv = – ∫ \(\frac{d x}{x}\)
⇒ – cos v = – log |x| + c
⇒ – cos \(\frac{y}{x}\) = – log |x| + c ……………….(2)
Since y(1) = 0 i.e. when x= 1 ; y = 0
∴ from (2); we have
– 1 = 0 + c
c = – 1
∴ From (2); we have
1 – cos \(\frac{y}{x}\) = – log |x|
cos \(\frac{y}{x}\) – 1= log | x | which is the required solution.

(vi) Given \(\frac{d y}{d x}=\frac{y}{x}-\tan \left(\frac{y}{x}\right)\) …………………..(1)
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\) in eqn. (1) ;
v + x \(\frac{d v}{d x}\) = v – tan v
⇒ x \(\frac{d v}{d x}\) = – tan v
⇒ \(\frac{d v}{\tan v}=-\frac{d x}{x}\)
On integrating both sides
\(\int \frac{d v}{\tan v}=-\int \frac{d x}{x}\) + log C
⇒ log |sin v| + log |x| = log C
⇒ x sin \(\frac{y}{x}\) = A …………………..(1)
[∵ A = ± C]
Given y = \(\frac{\pi}{4}\) at x = 1
∴ from (1) ;
\(\frac{1}{\sqrt{2}}\) = A
Thus eqn. (1) becomes ;
x tan \(\frac{y}{x}\) = \(\frac{1}{\sqrt{2}}\) be the required solution.

(vii) Given x e^y/x – y + x \(\frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=\frac{y-x e^{y / x}}{x}\)
\(\frac{y}{x}-e^{y / x}\) …………………(1)
which is homogeneous diff. eqn.
put y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ From (1) ; we have
v + x \(\frac{d v}{d x}\) = v – e^v
⇒ x \(\frac{d v}{d x}\) = – e^v
⇒ e^{– v} dv = – \(\frac{d x}{x}\) ;
on integrating we have
∫ e^{– v} dv = ∫ – \(\frac{d x}{x}\)
Since y (e) = 0
i.e. when x = e; y = 0
∴ From (2);
– 1 + log |e| = c
⇒ c = 0
∴ From (2) ;
e^{– y/x} = log |x|
⇒ \(-\frac{y}{x}\) = log (log |x|)
⇒ y = – x log (log |x|)
which is the required solution.

Question 17.
Solve the differential equation x dy – y dx = \(\sqrt{x^2+y^2}\) dx, given that y = 0 when x = 1.
Solution:
Given diff. eqn. can be written as ;
x dy = (y + \(\sqrt{x^2+y^2}\)) dx
⇒ \(\frac{d y}{d x}=\frac{y+\sqrt{x^2+y^2}}{x}\) ……………….(1)
putting y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
in eqn. (1) ; we have

Question 18.
Show that the following differential equations are homogeneous and find their particular solutions:
(i) x \(\frac{d y}{d x}\) sin (\(\frac{y}{x}\)) + x – y sin (\(\frac{y}{x}\)) = 0, given that x = 1 when y = \(\frac{\pi}{2}\).
(ii) (x e^y/x + y) dx = x dy, given that y = 0 when x = 1.
Solution:
(i) Given diff. eqn. be,

(ii) Given diff. eqn. be,
\(\frac{d y}{d x}=\frac{x e^{y / x}+y}{x}\) ;
putting y = vx
⇒ \(\frac{d y}{d x}\) = v + x \(\frac{d v}{d x}\)
∴ from (1) ; we get
– e v + x \(\frac{d v}{d x}\) = \(\frac{x e^v+v x}{x}\)
= e^v + v
⇒ x \(\frac{d v}{d x}\) = \(\frac{x e^v+v x}{x}\)
= e^v + v
⇒ x \(\frac{d v}{d x}\) = e^v
⇒ \(\int \frac{d v}{e^v}=\int \frac{d x}{x}\) + C
⇒ – e^-v = log |x| + C
⇒ – e^-y/x = log |x| + C ……………..(1)
When x = 1, y = 1
∴ from (1) ; we get
– e^-1 = c
c = – \(\frac{1}{e}\)
∴ from (1) ;
– e^-y/x = log |x| – \(\frac{1}{e}\)
which is the required solution.

Practicing ISC Maths Class 12 Solutions Chapter 9 Differential Equations Ex 9.5 is the ultimate need for students who intend to score good marks in examinations.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.5

Solve the following (1 to 5) differential equations:

Question 1.
(i) (x + y)² \(\frac{d y}{d x}\) = 1
(ii) \(\frac{d y}{d x}\) = (4x + y + 1)²
Solution:
(i) Given (x + y)² \(\frac{d y}{d x}\) = 1
putting x + y = t
⇒ 1 + \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\)
⇒ \(\frac{d y}{d x}\) = \(\frac{d y}{d x}\) – 1
∴ from eqn. (1) ; we have
t² [\(\frac{d y}{d x}\) – 1] = 1
⇒ \(\frac{d t}{d x}\) = \(\frac{1}{t^2}\) + 1
⇒ \(\frac{d t}{d x}=\frac{1+t^2}{t^2}\)
⇒ \(\frac{t^2 d t}{t^2+1}\) = dx
on integrating ; we have
∫ \(\left[\frac{t^2+1-1}{t^2+1}\right]\) dt = ∫ dx + c
⇒ ∫ [1 – \(\frac{1}{t^2+1}\)] dt = x + c
⇒ t – tan^-1 t = x c
⇒ x + y – tan^-1 (x + y) = x + c
⇒ y – tan^-1 (x + y) = c be the required solution.

(ii) Given eqn. is
\(\frac{d y}{d x}\) = (4x + y + 1)²
put 4x + y + 1 = t ;
Diff. w.r.t. x, we have
4 + \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\)
⇒ \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\) – 4
∴ From (1) ; we have
\(\frac{d t}{d x}\) – 4 = t²
⇒ \(\frac{d t}{d x}\) = t² + 4
⇒ \(\frac{d t}{t^2+4}\) = dx
[seperation of variables]
On integrating ; we have
\(\frac{1}{2}\) tan^-1 (\(\frac{t}{2}\)) = x + C
⇒ \(\frac{1}{2} \tan \left[\frac{4 x+y+1}{2}\right]\) = x + C is the required solution.

Question 2.
(i) (x – y)² \(\frac{d y}{d x}\) = a²
(ii) \(\frac{d y}{d x}\) = tan² (x + y).
Solution:
(i) Given Diff. eqn. is
(x – y)² \(\frac{d y}{d x}\) = a² ……………….(1)
put x – y =t ;
Diff. w.r.t. x, we get

(ii) Given, \(\frac{d y}{d x}\) = tan² (x + y) …………….(1)
put x + y = t
⇒ 1 + \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\)
⇒ \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\) – 1
∴ from (1) ;
\(\frac{d t}{d x}\) – 1 = tan² t
⇒ \(\frac{d t}{d x}\) = 1 + tan² t = sec² t
On variable seperation, we have
∫ \(\frac{d t}{\sec ^2 t}\) = ∫ dx + C
⇒ ∫ cos² t = ∫ dx + C
⇒ ∫ \(\left[\frac{1+\cos 2 t}{2}\right]\) dt = x + C
⇒ \(\frac{1}{2}\left[t+\frac{\sin 2 t}{2}\right]\) = x + C
⇒ x + y + \(\frac{1}{2}\) sin 2 (x + y) = 2x + A
⇒ y – x + \(\frac{1}{2}\) sin 2 (x + y) = A
⇒ 2 (y – x) + sin 2 (x + y) = 2A = A’
which is the required solution.

Question 3.
(i) \(\frac{d y}{d x}\) = (3x + y + 4)²
(ii) cos (x + y) dy = dx
Solution:
(i) Given \(\frac{d y}{d x}\) = (3x + y + 4)² ……………….(1)
put 3x + y + 4 = t
⇒ 3 + \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\)
[Diff. both sides w.r.t x]
⇒ \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\) – 3
∴ from (1) ;
\(\frac{d t}{d x}\) – 3 = t²
⇒ \(\frac{d t}{d x}\) = t² + 3
⇒ \(\frac{d t}{t^2+3}\) = dx
[after variable separation]
On integrating ; we have

(ii) Given eqn. is cos (x + y) dy = dx
⇒ cos (x + y) = \(\frac{d x}{d y}\)
⇒ \(\frac{d y}{d x}\) = sec (x + y) ……………….(1)
put x + y = t ;
Diff. w.r.t. x, we get
1 + \(\frac{d y}{d x}\) = cos (x + y)
(ii) cos = \(\frac{d t}{d x}\)
⇒ \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\) – 1,
putting in eqn. (1)
\(\frac{d t}{d x}\) – 1 = sec t
⇒ \(\frac{d t}{d x}\) = sec t + 1
⇒ \(\frac{d t}{1+\sec t}\) = dx ;
On integrating

Question 4.
(i) \(\frac{d y}{d x}\) = cos (x + y)
(ii) cos² (x – 2y) = 1 – 2 \(\frac{d y}{d x}\).
Solution:
(i) Given \(\frac{d y}{d x}\) = cos (x + y) ……………….(1)
put x + y = t
⇒ 1 + \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\)

(ii) Given cos² (x – 2y) = 1 – 2 \(\frac{d y}{d x}\)
put x – 2y = t
⇒ 1 – 2 \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\)
∴ From eqn. (1) ; we have
cos² t = \(\frac{d t}{d x}\)
⇒ dx = \(\frac{d t}{\cos ^2 t}\) ;
on integrating ; we have
∫ dx = ∫ sec² t dt
⇒ x = tan t + c
⇒ x = tan (x – 2y) + c be the required solution.

Question 5.
(i) (x + y + 1) \(\frac{d y}{d x}\) = 1
(ii) \(\frac{d y}{d x}=\frac{x+y-1}{x+y+1}\)
Solution:
(i) Given (x + y + 1) \(\frac{d y}{d x}\) = 1
⇒ \(\frac{d y}{d x}=\frac{1}{x+y+1}\) …………………(1)
put x + y + 1 = t
⇒ 1 + \(\frac{d y}{d x}=\frac{d t}{d x}\)
⇒ \(\frac{d y}{d x}=\frac{d t}{d x}\) – 1
Thus from eqn. (1) ; we have
\(\frac{d t}{d x}\) – 1 = \(\frac{1}{t}\)
\(\frac{d t}{d x}\) = 1 + \(\frac{1}{t}\)
= \(\frac{t + 1}{t}\)) dt = 2 dx
⇒ \(\frac{t d t}{t+1}\) = dx ;
on integrating
∫ \(\left[1-\frac{1}{t+1}\right]\) dt = ∫ dx
⇒ t – log (t + 1) = x + c
⇒ x + y + 1 – log (x + y + 2) = x + c
⇒ log (x + y + 2) = y + 1 -c
⇒ log (x + y + 2) = y + c’
⇒ x + y +2 = e^{y + c’}
⇒ x = Ae^y – y – 2 be the required solution.

(ii) Given \(\frac{d y}{d x}=\frac{x+y-1}{x+y+1}\) ………………….(1)
put x + y = t;
Diff., both sides w.r.t. x
1 + \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\)
∴ from eqn. (1) ;
\(\frac{d t}{d x}\) – 1 = \(\frac{t-1}{t+1}\)
⇒ \(\frac{d t}{d x}=\frac{t-1}{t+1}+1=\frac{2 t}{t+1}\)
⇒ (\(\frac{t + 1}{t}\)) dt = 2 dx
on integrating ; we get
⇒ ∫ (1 + \(\frac{1}{t}\)) dt = 2 ∫ dx + C
⇒ t + log |t| = 2x + C
⇒ x + y + log (x + y) = 2x + C
⇒ y – x + log |x + y| = C
which is the required solution.

Question 6.
Find a particular solution of the differential equation (x + y + 1)² dy = dx, given that y = 0 when x = – 1.
Solution:
Given diff. eqn. can be written as
(x + y + 1)² = 1 ……………..(1)
putting x + y + 1 = t
⇒ 1 + \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\)
[Diff. both sides w.r.t. x]
⇒ \(\frac{d y}{d x}\) = \(\frac{d t}{d x}\) – 1
∴ from (1) ;
t² (\(\frac{d t}{d x}\) – 1) = 1
⇒ \(\frac{d t}{d x}=\frac{1}{t^2}\) + 1
⇒ \(\frac{d t}{d x}=\frac{t^2+1}{t^2}\)
⇒ \(\frac{t^2 d t}{t^2+1}\) = dx
On integrating; we have
∫ \(\left[\frac{t^2+1-1}{t^2+1}\right]\) dt = ∫ dx + C
⇒ ∫ [1 – \(\frac{1}{t^2+1}\)] dt = x + C
⇒ t – tan^-1 t = x + C
⇒ x + y + 1 – tan^-1 (x + y + 1) = x + C
⇒ y + 1 – tan^-1 (x + y + 1) = C …………….(2)
which gives the general solution of given differential equation.
For particular solution of eqn. (1) ; we have
given that y = 0
when x = – 1
∴from (2) ; we get
0 + 1 – tan^-1 (- 1 + 0 + 1) C
⇒ C = 1
Thus eqn. (2) becomes ;
y + 1 – tan^-1 (x + y + 1) = 1
⇒ tan^-1 (x + y + 1) = y
⇒ x + y + 1 = tan y be the required solution.

Well-structured ISC Mathematics Class 12 Solutions Chapter 9 Differential Equations Ex 9.4 facilitate a deeper understanding of mathematical principles.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.4

Very short answer type questions (1 to 6) :

Solve the following (1 to 20) differential equations:

Question 1.
(i) (x² + 4) \(\frac{d y}{d x}\) = 1
(ii) \(\frac{d y}{d x}\) = y sin x
Solution:
(i) Given diff. eqn. be,
(x² + 4) \(\frac{d y}{d x}\) = 1
⇒ dy = \(\frac{d x}{x^2+4}\) [after variable separation]
On integrating ; we have
∫ dy = ∫ \(\frac{d x}{x^2+4}\)
⇒ y = \(\frac{1}{2}\) tan^-1 (\(\frac{x}{2}\)) + C, be the required solution.

(ii) Given diff. eqn. be,
\(\frac{d y}{d x}\) = y sin x
⇒ \(\frac{d y}{y}\) = sin x dx
[after variable separation]
On integrating ; we have
∫ \(\frac{d y}{y}\) = ∫ sin x dx + C
⇒ log |y| = – cos x + C which is the required solution.

Question 2.
(i) \(\frac{d y}{d x}=\sqrt{4-y^2}\) (NCERT)
(ii) \(\frac{d y}{d x}\) + y = 1. (NCERT)
Solution:
(i) Given, \(\frac{d y}{d x}=\sqrt{4-y^2}\)
⇒ \(\frac{d x}{d y}=\frac{1}{\sqrt{4-y^2}}\) ;
on integrating
x = ∫ \(\frac{d y}{\sqrt{2^2-y^2}}\) + C
⇒ x = sin^-1 (\(\frac{y}{2}\)) + C
⇒ x – C = sin^-1 \(\frac{y}{2}\)
⇒ y = 2 sin (x – C) is the required general solution.

(ii) Given diff. eqn. be,
\(\frac{d y}{d x}\) + y = 1
⇒ \(\frac{d y}{d x}\) = 1 – y
⇒ \(\frac{d y}{1-y}\) = dx
On integrating, we have
∫ \(\frac{d y}{1-y}\) = ∫ dx + C
⇒ \(\frac{\log |1-y|}{-1}\) = x + C
⇒ x + log |1 – y| = A be the required solution.

Question 3.
(i) x⁵ \(\frac{d y}{d x}\) = – y⁵ (NCERT)
(ii) x (1 + y²) dx + y (1 +x²) dy = 0
Solution:
(i) Given diff. eqn. be,
x⁵ \(\frac{d y}{d x}\) = – y⁵
⇒ \(\frac{1}{y^5} d y=-\frac{d x}{x^5}\) [variable separation]
On integrating ; we have
\(\int \frac{d y}{y^5}=-\int \frac{d x}{x^5}\) + C
⇒ \(-\frac{1}{4 y^4}=+\frac{1}{4 x^4}\) + C
⇒ \(\frac{1}{x^4}+\frac{1}{y^4}\) = A, which is the required soln.

(ii) Given diff. eqn. be
x (1 + y²) dx + y (1 + x²) dy = 0
On dividing throughout by
(1 + x²) (1 + y²) ; we have
\(\frac{x d x}{1+x^2}+\frac{y d y}{1+y^2}\) = 0
[after variable separation]
On integrating ; we have
\(\int \frac{x d x}{1+x^2}+\int \frac{y d y}{1+y^2}=\frac{1}{2} \log \mathrm{C}\)
⇒ \(\frac{1}{2} \int \frac{2 x d x}{1+x^2}+\frac{1}{2} \int \frac{2 y d y}{1+y^2}=\frac{1}{2} \log \mathrm{C}\)
⇒ \(\frac{1}{2}\) log (1 + x²) + \(\frac{1}{2}\) log (1 + y²) = \(\frac{1}{2}\) log C
⇒ log (1 + x²) (1 + y²) = log C
⇒ (1 + x²) (1 + y²) = C
which is the required solution.

Question 4.
(i) \(\frac{d y}{d x}=\frac{x+1}{2-y}\) (NCERT)
(ii) e^{y – x} \(\frac{d y}{d x}\) = 1
Solution:
(i) Given, \(\frac{d y}{d x}=\frac{x+1}{2-y}\)
⇒ (2 – y) dy = (x + 1) dx [variable separation]
On integrating ; we have
⇒ ∫ (2 – y) dy = ∫ (x + 1) dx + C
⇒ – \(\frac{1}{2}\) (2 – y)² = \(\frac{(x+1)^2}{2}+\frac{C}{2}\)
⇒ x² + 2x + 1 + y² – 4y + C + 4 = 0
⇒ x² + y² + 2x – 4y + C’ = 0
is the required solution.

(ii) Given e^{y – x} \(\frac{d y}{d x}\) = 1
⇒ e^{+ y} dy = e^x dx
On integrating both sides ; we have
∫ e^y dy = ∫ e^x dx + C
⇒ e^y = e^x + C
which is the required solution.

Question 4 (old).
(ii) \(\frac{d y}{d x}\) = e^{x + y} (NCERT)
Solution:
Given, \(\frac{d y}{d x}\) = e^{x + y}
⇒ \(\frac{d y}{e^y}\) = e^x dx
[after variable separation]
⇒ ∫ e^{– y} dy = ∫ e^x dx + C
⇒ – e^{– y} = e^x + C which is the required solution.

Question 5.
(i) \(\frac{d y}{d x}\) = (e^x + 1) y
(ii) \(\frac{d y}{d x}+\frac{1+y^2}{y}\) = 0
Solution:
(i) Given, \(\frac{d y}{d x}\) = (e^x + 1) y
⇒ \(\frac{d y}{y}\) = (e^x + 1) dx
[after variable separation]
∫ \(\frac{d y}{y}\) = ∫ (e^x + 1) dx + C
⇒ log |y| = e^x + x + C
which is the required solution.

(ii) Given diff. eqn. be,
\(\frac{d y}{d x}+\frac{1+y^2}{y}\) = 0
⇒ \(\frac{d y}{d x}=-\frac{\left(1+y^2\right)}{y}\)
⇒ \(\frac{y d y}{1+y^2}\) = – dx + C
On integrating both sides, we have
⇒ ∫ \(\frac{y d y}{1+y^2}\) = ∫ – dx + C
put 1 + y² = t
⇒ 2y dy = dt
⇒ ∫ \(\frac{d t}{2 t}\) = – x + c
⇒ \(\frac{1}{2}\) log |t| + x = c
⇒ \(\frac{1}{2}\) log |1 + y²| + x = c be the reqd. solution.

Question 6.
(i) \(\frac{d y}{d x}=\frac{1+y^2}{1+x^2}\)
(ii) \(\frac{d y}{d x}\) = 2^{y – x} (NCERT Exemplar)
Solution:
(i) Given, \(\frac{d y}{d x}=\frac{1+y^2}{1+x^2}\)
⇒ \(\frac{1}{1+y^2} d y=\frac{1}{1+x^2} d x\)
[after variable separation]
On integrating ; we have
\(\int \frac{d y}{1+y^2}=\int \frac{d x}{1+x^2}\) + c
⇒ tan^-1 y = tan^-1 x + c, be the required soln.

(ii) Given, \(\frac{d y}{d x}\) = 2^{y – x}
⇒ 2^-y dy = 2^-x dx
[after variable separation]
On integrating; we have
∫ 2^-y dy = ∫ 2^-x dx + C
⇒ \(\frac{2^{-y}}{-\log 2}=\frac{2^{-x}}{-\log 2}\) + C
⇒ 2^-y = 2^-x – C log 2
⇒ 2^-y – 2^-x = A, be the required solution.

Question 7.
(i) \(\frac{d y}{d x}\) = log x
(ii) sin (\(\frac{d y}{d x}\)) = a.
Solution:
(i) Given diff. eqn. be \(\frac{d y}{d x}\) = log x
after variable separation, we have
dy = log x dx
On integrating both sides, we have
∫ dy = ∫ log x . 1 dx
⇒ y = (log x) x – ∫ \(\frac{1}{x}\) . x dx
⇒ y = x log x – x + c
be the required solution of given differential equation.

(ii) Given diff. eqn. be
sin (\(\frac{d y}{d x}\)) = a.
⇒ \(\frac{d y}{d x}\) = sin^-1 a
On integrating both sides ; we have
y = (sin^-1 a) x + C
which is the required solution.

Question 7 (old).
(ii) (sin⁴ x) \(\frac{d y}{d x}\) = cos x
Solution:
Given diff eqn. be (sin⁴ x) \(\frac{d y}{d x}\) = cos x
after variable separation. we have
dy = \(\frac{\cos x}{\sin ^4 x}\) dx
On integrating both sides; we have
∫ dy = ∫ \(\frac{\cos x d x}{\sin ^4 x}\) dx
put sin x = t
⇒ cos x dx = dt
∫ dy = ∫ \(\frac{d t}{t^4}\) + C
⇒ y = \(\frac{t^{-3}}{-3}\) + C
c y = – \(\frac{1}{3}\) cosec³ x + c be the required solution.

Question 8.
(i) e^x \(\frac{d y}{d x}\) + 1 = x
(ii) \(\frac{1}{x} \cdot \frac{d y}{d x}\) = tan^-1 x
Solution:
(i) Given, e^x \(\frac{d y}{d x}\) + 1 = x
⇒ dy = \(\frac{(x-1)}{e^x}\) dx
On integrating ; we have
∫ dy = ∫ (x – 1) e^-x dx + C
⇒ y = (x – 1) \(\frac{e^{-x}}{-1}\) + ∫ 1 . e^-x dx + C
⇒ y = – (x – 1) e^-x – e^-x + C
⇒ y = – x e^-x + C, which is the required solution.

(ii) Given diff. eqn. be,
\(\frac{1}{x} \frac{d y}{d x}\) = tan^-1 x – x, x ≠ 0
after variable separation, we have
dy = (tan^-1 x) x dx
On integrating both sides, we have
∫ dy = ∫ x tan^-1 x dx
⇒ y = \(\tan ^{-1} x \cdot \frac{x^2}{2}-\int \frac{1}{1+x^2} \frac{x^2}{2} d x\)
⇒ y = \(\frac{x^2}{2} \tan ^{-1} x-\frac{1}{2} \int\left[\frac{1+x^2-1}{1+x^2}\right] d x\)
⇒ y = \(\frac{x^2}{x} \tan ^{-1} x-\frac{1}{2} \int\left[1-\frac{1}{1+x^2}\right] d x\)
⇒ y = \(\frac{x^2}{2} \tan ^{-1} x-\frac{x}{2}+\frac{1}{2} \tan ^{-1} x\) + c
i.e. y = \(\frac{1}{2}\) (x² + 1) tan^-1 x – \(\frac{x}{2}\) + c be the reqd. solution.

Question 9.
(i) y’ = (cos² x – sin² x) cos² y
(ii) (xy² + x) dx + (x²y + y) dy = 0 (ISC 2012)
(iii) (x² – yx²) dy + (y² + xy²) dx = 0
Solution:
(i) Given \(\frac{d y}{d x}\) = (cos² x – sin² x) cos² y
⇒ \(\frac{1}{\cos ^2 y}\) dy = cos 2x dx
on integrating both sides ; we have
∫ sec² y dy = ∫ cos 2x dx
⇒ tan y = \(\frac{\sin 2 x}{2}\) + c ;
which is the required solution.

(ii) Given, (xy² + x) dx + (x²y + y) dy = 0
⇒ x (y² + 1) dx + y (x² + 1) dy = 0
On dividing throughout the eqn. (1) by (x² + 1) (y² + 1); we have
\(\frac{x d x}{x^2+1}+\frac{y d y}{y^2+1}\) = 0
On integrating; we have
\(\int \frac{x d x}{x^2+1}+\int \frac{y d y}{y^2+1}\) = \(\frac{1}{2}\) log C
⇒ \(\frac{1}{2}\) log(1 + x²) + \(\frac{1}{2}\) log (1 + y²) = \(\frac{1}{2}\) log C
⇒ log (1 + x²) (1 + y²) = log C
⇒ (1 + x²) (1 + y²) = C be the required soln.

(iii) Given differential eqn. be,
(x² – yx²) dy + (y² + xy²) dx = 0
⇒ x² (1 – y) dy + y² (1 + x) dx = 0
Dividing throughout by x²y² ; we have
\(\frac{-(1-y)}{y^2} d y+\frac{(1+x)}{x^2} d x\) = 0
On integrating both sides ; we have
\(\int \frac{1}{y^2} d y-\int \frac{1}{y} d y+\int \frac{d x}{x^2}+\int \frac{d x}{x}\) = C
⇒ \(-\frac{1}{y}-\log |y|+\left(-\frac{1}{x}\right)+\log |x|\) = C
⇒ \(\log \left|\frac{x}{y}\right|-\frac{1}{x}-\frac{1}{y}\) = C be the required soln.

Question 10.
(i) (1 + x) (1 + y²) dx + (1 + y) (1 + x²) dy = 0
(ii) x \(\sqrt{1-y^2}\) dx + y \(\sqrt{1-x^2}\) dy = 0
Solution:
(i) Given, (1 + x) (1 + y²) dx + (1 + y) (1 + x²) dy = 0
Dividing throughout given diff. eqn. by (1 + x²) (1 + y²) ; we have
⇒ \(\frac{(1+x)}{1+x^2} d x+\frac{(1+y)}{1+y^2} d y\) = 0
[after variable separation]
On integrating ; we have

(ii) Given, x \(\sqrt{1-y^2}\) dx + y \(\sqrt{1-x^2}\) dy = 0 ;
On dividing throughout by \(\sqrt{1-x^2} \sqrt{1-y^2}\) ; we get

Question 11.
(i) cos x cos y dy + sin x sin y dx = 0
(ii) cos x (1 + cos y) dx – sin y (1 + sin x) dy = 0
Solution:
(i) Given cos x cos y \(\frac{d y}{d x}\) = – sin x sin y ;
after variable seperation, we have
\(\frac{\cos y d y}{\sin y}=\frac{-\sin x d x}{\cos x}\)
on integrating both sides ; we have
\(\int \frac{\cos y d y}{\sin y}=\int \frac{-\sin x d x}{\cos x}\) + log c
⇒ log sin y = log cos x + log c
⇒ sin y = c cos x be the required solution.

(ii) Given, cos x (1 + cos y) dx – sin y (1 + sin x) dy = 0
⇒ \(\frac{\cos x d x}{1+\sin x}-\frac{\sin y d y}{1+\cos y}\) = 0
[after variable seperation]
On integrating both sides ; we have
\(\int \frac{\cos x d x}{1+\sin x}+\int \frac{-\sin y d y}{1+\cos y}\) = C
⇒ log (1 + sin x) (1 + cos y) = log A
⇒ (1 + sin x) (1 + cos y) = A be the required soln.

Question 12.
(i) x (e^2y – 1) dy + (x² – 1) e^y dx = 0
(ii) e^x tan y + (1 – e^x) sec² y dy = 0
Solution:
(i) Given, x (e^2y – 1) dy + (x² – 1) e^y dx = 0
⇒ \(\frac{\left(e^{2 y}-1\right)}{e^y} d y+\frac{\left(x^2-1\right)}{x} d x\) = 0
[after variable separation]
On integrating ; we have
∫ (e^y – e^{– y}) dy + ∫ (x – \(\frac{1}{x}\)) dx = C
⇒ e^y + e^{– y} + \(\frac{x^2}{2}\) – log |x| + C
be the required solution.

(ii) Given e^x tan y + (1 – e^x) sec² y dy = 0
Dividing throughout by (1 – e^x) tan y ; we have
\(\frac{e^x}{1-e^x}+\frac{\sec ^2 y d y}{\tan y}\) = 0
On integrating ; we have
\(\int \frac{e^x d x}{1-e^x}+\int \frac{\sec ^2 y d y}{\tan y}\) = log C
⇒ – \(\int \frac{-e^x d x}{1-e^x}+\int \frac{\sec ^2 y d y}{\tan y}\) = log C
⇒ – log |1 – e^x| + log |tan y| = log C
⇒ log \(\frac{\tan y}{1-e^x}\) = log C
⇒ tan y = C (1 – e^x), which is required solution.

Question 13.
(i) y dx – x dy = xy dx
(ii) (1 + y²) tan^{– 1} x dx + 2y (1 + x²) dy = 0 (NCERT Exemplar)
Solution:
(i) Given, y dx – x dy = xy dx
⇒ (y – xy) dx – x dy = 0
⇒ y (1 – x) dx – x dy = 0
⇒ \(\left(\frac{1-x}{x}\right) d x-\frac{d y}{y}\) = 0 ;
on integrating
\(\int\left(\frac{1}{x}-1\right) d x-\int \frac{d y}{y}\) = C
log |x| – x – log |y| = C
⇒ log |y| – log |x| = – x – C
⇒ log \(\left|\frac{y}{x}\right|\) = – x – c
⇒ y = Axe^-x
[Here A = ± e^-c]
which is the required soln.

(ii) Given, (1 + y²) tan^-1 x dx + 2y (1 + x²) dy = 0
⇒ \(\frac{\tan ^{-1} x d x}{1+x^2}+\frac{2 y d y}{1+y^2}\)= 0
[after variable separation]
On integrating ; we have
⇒ \(\frac{\left(\tan ^{-1} x\right)^2}{2}\) + log (1 + y²) = C,
be the required soln.
[∵ ∫ [f(x)]ⁿ f'(x) dx = \(\frac{[f(x)]^{n+1}}{n+1}\), n ≠ – 1
and ∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)|]

Question 14.
(i) (1 + x²) dy = x y dx
(ii) xyy’ = 1 + x + y + xy
Solution:
(i) Given, (1 + x²) dy = x y dx
⇒ \(\frac{d y}{y}=\frac{x d x}{1+x^2}\)
[after variable seperation]
On integrating ; we have
⇒ \(\int \frac{d y}{y}=\frac{1}{2} \int \frac{2 x d x}{1+x^2}\) + log C
⇒ log y = \(\frac{1}{2}\) log (1 + x²) + log C
⇒ y = C \(\sqrt{1+x^2}\) which is the required solution.

(ii) xyy’ = 1 + x + y + xy
⇒ \(\frac{y d y}{1+y}=\frac{1+x}{x} d x\)
[after variable seperation]
On integrating ; we have
\(\int\left[1-\frac{1}{1+y}\right] d y=\int\left(\frac{1}{x}+1\right) d x\)
y – log |1 + y| = log |x| + x + C be the required solution.

Question 15.
(i) y (1 – x²) \(\frac{d y}{d x}\) = x (1 + y²)
(ii) (y + xy) dx + x (1 – y²) dy = 0
Solution:
(i) Given, y (1 – x²) \(\frac{d y}{d x}\) = x (1 + y²)
after variable separation, we have
\(\frac{y d y}{1+y^2}=\frac{x d x}{1-x^2}\)
On integrating both sides, we have
\(\frac{1}{2} \int \frac{2 y d y}{1+y^2}=\frac{-1}{2} \int \frac{-2 x d x}{1-x^2}\)
⇒ \(\frac{1}{2}\) log |1 + y²| + \(\frac{1}{2}\) log |1 – x²| = \(\frac{1}{2}\) log c
⇒ \(\frac{1}{2}\) log |(1 + y²) (1 – x²) = \(\frac{1}{2}\) log c
⇒ (1 + y²) (1 – x²) = c be the required solution.

(ii) Given, (y + xy) dx + (x – xy²) dy = 0
⇒ y (1 + x) dx + x (1 – y²) dy = 0
⇒ \(\frac{(1+x)}{x}\) dx + \(\frac{\left(1-y^2\right)}{y}\) dy = 0
On integrating both sides, we have
\(\int\left[\frac{1}{x}+1\right] d x+\int\left[\frac{1}{y}-y\right] d y\) = 0
⇒ log |x| + x + log |y| – \(\frac{y^2}{2}\) = c
⇒ log |xy| + x – \(\frac{y^2}{2}\) = C
which is the required solution.

Question 16.
(i) (x² – yx²) dy + (y² + x²y²) dx = 0
(ii) (e^x + 1) y dy = (y + 1) e^x dx
Solution:
(i) Given, (x² – yx²) dy + (y² + x²y²) dx = 0
⇒ x² (1 – y) dy + y² (1 + x²) dx = 0
⇒ \(\frac{(1-y) d y}{y^2}+\frac{\left(1+x^2\right)}{x^2} d x\) = 0
On integrating ; we have
\(\int\left[\frac{1}{y^2}-\frac{1}{y}\right] d y+\int\left(\frac{1}{x^2}+1\right) d x\) = 0
⇒ – \(\frac{1}{y}\) – log |y| – \(\frac{1}{x}\) + x = – C
⇒ log |y| + \(\frac{1}{y}\) + \(\frac{1}{x}\) – x = C,
which is the required solution.

(ii) Given, y (1 + e^y) dy = (y + 1) e^x dx
after variable separation, we have
\(\int \frac{y d y}{1+y}=\int \frac{e^x d x}{1+e^x}\)
⇒ \(\int\left(1-\frac{1}{1+y}\right) d y=\int \frac{e^x d x}{e^x+1}\)
⇒ y – log |1 + y| = log |e^x + 1| + c
y = log |(e^x + 1) (1 + y)| + C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(x)}\) dx = log |f(x)|]
which is the required solution.

Question 17.
(i) \(\frac{d y}{d x}\) = e + x² e^{– 2y}
(ii) (1 + e^2x) dy + (1 + y²) e^x dx = 0
Solution:
(i) Given, \(\frac{d y}{d x}\) = e + x² e^{– 2y}
⇒ \(\frac{d y}{d x}\) = e^{– 2y} (e^3x + x²)
⇒ e^2y dy = (e^3x + x²) dx
On integrating ; we have
∫ e^2y dy = ∫ (e^3x + x²) dx
\(\frac{e^{2 y}}{2}=\frac{e^{3 x}}{3}+\frac{x^3}{3}\) + C,
which is the required solution.

(ii) Given, (1 + e^2x) dy + (1 + y²) e^x dx = 0
⇒ \(\frac{d y}{1+y^2}+\frac{e^x d x}{1+e^{2 x}}\) = 0
On integrating ; we have
\(\int \frac{d y}{1+y^2}+\int \frac{e^x d x}{1+e^{2 x}}\) = C
putting e^x = t
⇒ e^x dx = dt in 2nd integral
⇒ tan^-1 y + ∫ \(\frac{d t}{1+t^2}\) = C
⇒ tan^-1 y + tan^-1 (e^x) = C
which is the required solution.

Question 18.
(i) \(\sqrt{a+x} \frac{d y}{d x}\) + xy = 0
(ii) (x – 1) dy = 2x³y dx.
Solution:
(i) Given, \(\sqrt{a+x} \frac{d y}{d x}\) + xy = 0
⇒ \(\frac{d y}{y}+\frac{x d x}{\sqrt{a+x}}\) = 0
On integrating ; we have
⇒ \(\int \frac{d y}{y}+\int \frac{a+x-a}{\sqrt{a+x}} d x\) = C
⇒ log |y| + ∫ \(\left[\sqrt{a+x}-\frac{a}{\sqrt{a+x}}\right]\) dx = C
⇒ log |y| + \(\frac{2}{3}\) (a + x)^3/2 – 2a \(\sqrt{a + x}\) = C
⇒ log |y| + \(\frac{2}{3}\) \(\sqrt{a + x}\) (a + x – 3a) = C
⇒ log |y| = – \(\frac{2}{3}\) (x – 2a) \(\sqrt{a + x}\) + C
which is the required solution.

(ii) Given, (x – 1) dy = 2x³ y dx
\(\frac{d y}{y}=\frac{2 x^3}{x-1} d x\) ;
On integrating
\(\int \frac{d y}{y}=2 \int\left(\frac{x^3-1+1}{x-1}\right)\) dx
log |y| = 2 ∫ [x² + x + 1) + \(\frac{1}{x-1}\)] dx + C
⇒ log |y| = 2 [\(\frac{x^3}{3}+\frac{x^2}{2}\) + x + log |x – 1|] + C
which is the required solution.

Question 19.
(i) x log x dy – y dx = 0
(ii) sin³ x dx – sin y dy = 0
Solution:
(i) Given, x log x dy – y dx = 0
⇒ \(\frac{d y}{y}-\frac{d x}{x \log x}\) = 0
[after variable seperation]
On integrating ; we have
\(\int \frac{d y}{y}-\int \frac{\frac{1}{x} d x}{\log x}\) = log A
⇒ log y – log (log x) = log A
⇒ log y = log (A log x)
⇒ y = A log x, be the required solution.

(ii) Given, sin³ x dx – sin y dy = 0
On integrating ; we have
∫ sin³ x dx = ∫ sin y dy + C
⇒ ∫ (1 – cos² x) sin x dx = – cos y dy + C
put cos x = t
⇒ – sin x dx = dt
⇒ (1 – t²) (- dt) = – cos y + C
⇒ – [t – \(\frac{t^3}{3}\)]= – cos y + c
⇒ cos y = cos x – \(\frac{\cos ^3 x}{3}\) + C
which is the required solution.

Question 20.
(i) e^{2x – 3y} dx + e^{2y – 3x} dy = 0
(ii) log (\(\frac{d y}{d x}\)) = 2x – 3y (ISC 2013)
Solution:
(i) Given, e^{2x – 3y} dx + e^{2y – 3x} dy = 0
⇒ e^2x e^{– 3y} dx + e^2y e^{– 3x} dy = 0
⇒ \(\frac{e^{2 x}}{e^{-3 x}} d x+\frac{e^{2 y}}{e^{-3 y}} d y\) = 0
⇒ e^5x dx + e^5y dy = 0
On integrating ; we have
⇒ ∫ e^5x dx + ∫ e^5y dy = \(\frac{C}{5}\)
⇒ \(\frac{e^{5 x}}{5}+\frac{e^{5 y}}{5}=\frac{C}{5}\)
⇒ e^5x + e^5y = C, be the required solution.

(ii) Given, log (\(\frac{d y}{d x}\)) = 2x – 3y
⇒ \(\frac{d y}{d x}\) = e^{2x – 3y}
⇒ e^3y dy = e^2x dx
On integrating ; we have
∫ e^3y dy = ∫ e^2x dx
\(\frac{e^{3 y}}{3}=\frac{e^{2 x}}{2}\) + C
which is the required solution.

Question 21.
If \(\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}\) = 0, show that \(\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}\) = A.
Solution:
Given, \(\frac{d y}{d x}+\sqrt{\frac{1-y^2}{1-x^2}}\) = 0
⇒ \(\frac{d y}{\sqrt{1-y^2}}+\frac{d x}{\sqrt{1-x^2}}\) = 0
[after variable seperation]
On integrating ; we have
sin^-1 y + sin^-1 x = C
⇒ sin^-1 [y \(\sqrt{1-x^2}\) + x \(\sqrt{1-y^2}\)] = C
⇒ y \(\sqrt{1-x^2}\) + x \(\sqrt{1-y^2}\) = sin C = A.

Question 22.
Solve the differential equation \(\frac{d y}{d x}\) = 1 + x + y² + xy², given that y = 0 when x = 0. (NCERT Exemplar)
Solution:
Given diff. eqn. be,
\(\frac{d y}{d x}\) = 1 + x + y² + xy²
⇒ \(\frac{d y}{d x}\) = (1 + x) (1 + y²)
⇒ \(\frac{d y}{1+y^2}\) = (1 + x) dx
On integrating ; we have
∫ \(\frac{d y}{1+y^2}\) = ∫ (1 + x) dx + C
⇒ tan^-1 y = x + \(\frac{x^2}{2}\) + C …………………(1)
Since y = 0, when x = 0
∴ from (1);
0 = 0 + 0 + C
⇒ C = 0
Thus eqn. (1) gives ;
tan^-1 y = x + \(\frac{x^2}{2}\)
⇒ y = tan (x + \(\frac{x^2}{2}\))
which is the required solution.

Question 23.
Find the particular solution of the differential equation \(\frac{d y}{d x}\) = – 4xy², given that y = 1 when x = 0. (NCERT)
Solution:
Given, \(\frac{d y}{d x}\) = – 4xy²
⇒ \(\frac{d y}{y^2}\) = – 4x dx
[variable separation]
On integrating; we have
∫ \(\frac{d y}{y^2}\) = ∫ – 4x dx + C
⇒ – \(\frac{1}{y}\) = – 2x² + C …………….(1)
y(0) = 1
⇒ When x = 0, y = 1
∴ – 1 = C
∴ from (1) ;
– \(\frac{1}{y}\) = – 2x² – 1
∴ y = \(\frac{1}{2 x^2+1}\) is the required solution.

Question 24.
Solve the following differential equations:
(i) 2 (y + 3) – xy \(\frac{d y}{d x}\) = 0, given that y(1) = – 2. (NCERT Exemplar)
(ii) (1 + y²) dx + x dy = 0, given that y (1) = 1.
(iii) \(\frac{d y}{d x}\) = x² e^-3y, given that y = 0 for x = 0.
Solution:
(i) Given 2 (y + 3) – xy \(\frac{d y}{d x}\) = 0
⇒ 2 \(2 \frac{d x}{x}-\frac{y d y}{y+3}\) = 0
On integrating both sides ; we have
\(2 \int \frac{d x}{x}-\int\left(\frac{y+3-3}{y+3}\right) d y\) = 0
⇒ 2 log x – ∫ \(\left[1-\frac{3}{y+3}\right]\) dy = 0
⇒ 2 log x – y + 3 log |y + 3| = c ……………….(1)
Since y(1) = – 2
i.e. when x = 1 ; y = 2
∴ from (1) ; we have
2 log 1 – (- 2) + 3 log |1| = c
⇒ c = 2
∴ From (1) ; we have
2 log x – y + 3 log (y + 3) = 2
⇒ log x² (y + 3)³ = y + 2
⇒ x² (y + 3)³ = e^{y + 2} be the required solution.

(ii) Given (1 + y²) dx + xdy = 0
⇒ \(\frac{d x}{x}+\frac{d y}{1+y^2}\) = 0
[after variable separation]
On integrating ; we have
\(\int \frac{d x}{x}+\int \frac{d y}{1+y^2}\) = C
⇒ log |x| + tan^-1 y = C ………………(1)
Given y(1) = 1 i.e. When x = 1, y = 1
∴ from (1) ;
\(\frac{\pi}{4}\) = C
∴ eqn. (1) becomes,
log x + tan^-1 y = \(\frac{\pi}{4}\)
is the required solution.

(iii) Given \(\frac{d y}{d x}\) = x² e^{– 3y} ;
after separation of variables, we have
\(\frac{d y}{e^{-3 y}}\) = x² dx ;
On integrating we have
∫ e^3y dy = ∫ x² dx + C
⇒ \(\frac{e^{3 y}}{3}=\frac{x^3}{3}\) + C ……………….(1)
When x = 0, y = 0
∴ (1) gives ; \(\frac{1}{3}\) = C
∴ From (1) ;
e^3y = x³ + 1 is the required solution.

Question 25.
Solve the following differential equations:
(i) (1 + y²) (1 + log x) dx + x dy = 0, given that when x = 1, y = 1.
(ii) (1 + x²) \(\frac{d y}{d x}\) + (1 + y²) = 0, given that y = 1 when x = 0.
(iii) x (1 + y²) dx – y (1 + x²) dy = 0, given that y = 0 when x = 1.
Solution:
(i) Given, (1 + y) (1 + log x) dx + x dy = 0
⇒ \(\left(\frac{1+\log x}{x}\right) d x+\frac{d y}{1+y^2}\) = 0
[variable separation]
⇒ ∫ \(\frac{1}{x}\) dx + ∫ log x . \(\frac{1}{x}\) dx + tan^-1 y = C
⇒ log x + \(\frac{(\log x)^2}{2}\) + tan^-1 y = C
Given, when x = 1, y = 1
∴ eqn. (1) gives ;
\(\frac{\pi}{4}\) = C
∴ eqn. (1) becomes ;
log x + \(\frac{(\log x)^2}{2}\) + tan^-1 y = \(\frac{\pi}{4}\)
is the required solution.

(ii) Given (1 + x²) \(\frac{d y}{d x}\) + (1 + y²) = 0
⇒ \(\frac{d y}{1+y^2}+\frac{d x}{1+x^2}\) = 0
On integrating both sides ; we have
\(\int \frac{d y}{1+y^2}+\int \frac{d x}{1+x^2}\) = 0
⇒ tan^-1 x + tan^-1 y = tan^-1 c
⇒ tan^-1 \(\left[\frac{x+y}{1-x y}\right]\) = tan^-1 c
⇒ x + y = c (1 – xy) ……………..(1)
Given y = 1 when x = 0
∴ from (1) ; we have
1 = c (1 – 0)
⇒ c = 1.
∴ From (1) ; we get
x + y = 1 – xy be the required solution.

(iii) Given, x (1 + y²) dx – y (1 + x²) dy = 0
⇒ \(\frac{x d x}{1+x^2}-\frac{y d y}{1+y^2}\) = 0
On integrating ; we have
\(\frac{1}{2} \int \frac{2 x d x}{1+x^2}-\frac{1}{2} \int \frac{2 y d y}{1+y^2}\) = \(\frac{1}{2}\) log C
⇒ log (1 + x²) – log (1 +y²) = log C
[∵∫ \(\frac{f^{\prime}(x)}{f(x)}\) dx = log |f(x)|]
⇒ log \(\left(\frac{1+x^2}{1+y^2}\right)\) = log C
⇒ 1 + x² = C (1 + y²) ……………….(1)
given that y = 0 when x = 1
∴ from (1) ; 2 = C
Thus from (1) ; we get
(1 + x²) = 2 (1 + y²)
⇒ 2y² = x² – 1
which is the required soln.

Question 26.
Solve the following differential equations:
(i) y’ = y tan x, given that y = 1 when x = 0.
(ii) \(\frac{d y}{d x}\) = y² tan 2x, given that y (0) = 2
(iii) y’ = y cot 2x, given that y (\(\frac{\pi}{4}\)) = 2.
Solution:
(i) Given \(\frac{d y}{d x}\) = y tan x
⇒ \(\frac{d y}{y}\) = tan x dx
on integrating both sides ; we have
∫ \(\frac{d y}{y}\) = tan x dx
⇒ log |y| = – log |cos x| + log c
⇒ log (y cos x) = log c
⇒ y = cosx = c ………………(1)
Since y (0) = 1 i.e. when x = 0 ; y = 1
∴ from (1) ; we have
1 × cos 0 = c
⇒ c = 1.
Thus from eqn (1) ; we have
y cos x = 1
⇒ y = \(\frac{1}{cos x}\) = sec x be the reqd. solution.

(ii) Given \(\frac{d y}{d x}\) = y² tan 2x
after separation of variables
\(\frac{d y}{y^2}\) = tan 2x . dx;
On integrating; we get
∫ \(\frac{d y}{y^2}\) = ∫ tan 2x dx + C
⇒ \(-\frac{1}{y}=-\frac{\log |\cos 2 x|}{2}\) + C ………………(1)
given y (0) = 1
i.e. when x = 0, y = 1
∴ from (1) ;
– 1 = c
∴ eqn. (1) gives ;
\(-\frac{1}{y}=-\frac{1}{2}\) log |cos 2x| – 1
⇒ \(\frac{1}{y}=\frac{1}{2}\) log |cos 2x| + 1 is the required solution.

(iii) Given y’ = y cot 2x
⇒ \(\frac{d y}{y}\) = cot 2x dx ;
on integrating ; we get
\(\int \frac{d y}{y}=\int \frac{\cos 2 x}{\sin 2 x} d x+\frac{1}{2} \log c\)
⇒ log |y| = \(\frac{\log |\sin 2 x|}{2}+\frac{1}{2} \log c\)
⇒ y² = c sin 2x …………………(1)
Given, y (\(\frac{\pi}{4}\)) = 2
i.e. y = 2 when x = \(\frac{\pi}{4}\)
∴ 4 = c . 1
⇒ c = 4
∴ From (1) ;
y² = 4 sin 2x is the required solution.

Question 27.
Find the particular solution of cos (\(\frac{d y}{d x}\)) = a, given that y = 2 when x = 0. (NCERT)
Solution:
Given cos(\(\frac{d y}{d x}\) = a
⇒ \(\frac{d y}{d x}\) = cos^-1 a
⇒ dy = cos^-1 a dx ;
On integrating
∫ dy = ∫ cos^-1 a dx + C
⇒ y = x cos^-1 a + C ………………….(1)
Given y = 2 when x = 0
∴ from (1) ;
2 = 0 + C
⇒ C = 2
Thus eqn. (1) becomes ;
y = x cos^-1 a + 2
⇒ \(\frac{y-2}{x}\) = cos^-1 a
⇒ cos (\(\frac{y-2}{x}\)) = a, which is the required solution.