Raju

Practicing ISC Maths Class 12 Solutions 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 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.

Students can track their progress and improvement through regular use of Understanding ISC Mathematics Class 12 Solutions.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.3

Very short answer type questions (1 to 4) :

Question 1.
(i) Write the order of the differential of the family of circles x² + y² = a², where a (> 0) is arbitrary constant.
(ii) Write the order of the differential equation of the family of parabolas y² = 4ax, where a is arbitrary constant.
(iii) Write the order of the differential equation of the family of parabolas y² = 4a (x – b), a, b are arbitrary constants.
(iv) Write the order of the differential equation of the family of ellipses \(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1, a, b are arbitrary constants.
Solution:
(i) Given eqn. of family of circles be
x² + y² = a² ………………..(1)
where a be an arbitrary constant
duff. eqn. (1) w.r.t. x; we have
2x + 2yy’ = 0 ………………..(2)
The highest ordered derivative existing in eqn.(2) be \(\frac{d y}{d x}\) so its order be 1.

(ii) Given eqn. of family of parabolas be
y² = 4ax …………………..(1)
duff. eqn. (1) w.r.t. x, we have
2yy’ = 4a
The highest ordered derivative in given diff. eqn. be y’.
∴ its order be 1.

(iii) Given y² = 4a (x – b) ;
where a, b are arbitrary constants.
2yy’ = 4a ;
again diff. both sides w.r.t. x
2[yy” + y’²] = 0
⇒ yy” + y’² = 0
The highest ordered derivative existing in given diff. eqn. be y” so its order be 2.

(iv) Given eqn. of family of ellipses be;
\(\frac{x^2}{a^2}+\frac{y^2}{b^2}\) = 1 ……………(1)
Diff. eqn. (1) w.r.t. x ; we have

Here, the highest ordered derivative existing in diff. eq. (2) be \(\frac{d^2 y}{d x^2}\) and hence its order be 2.

Question 2.
(i) Form the differential equation of the . family of straight lines y = mx, where m is arbitrary constant.
(ii) Find the differential equation of the family of straight lines y = mx + c, where m, c are arbitrary constants.
Solution:
(i) Given y = mx ………………(1)
Diff. (1) w.r.t. x ; we have \(\frac{d y}{d x}\) = m ………………………(2)
Eliminating m from (1) and (2); we hae dy
y = x \(\frac{d y}{d x}\) is the required diff. eqn.

(ii) Given y = mx + c, where m, c are arbitrary constants
Differentiating w.r.t. x, we have dy
\(\frac{d y}{d x}\) = m ; again differentiating w.r.t. x,
we have \(\frac{d^2 y}{d x^2}\) = 0, be the required equation.

Question 3.
Form the differential equation of the family of concentric circles x² + y² = r², where r (> 0) is arbitrary constant.
Solution:
Given x² + y² = r² ……………(1)
Diff. both sides eqn. (1) w.r.t. x, we have
2x + 2y \(\frac{d y}{d x}\) = 0
=> x + y \(\frac{d y}{d x}\) = 0 be the required differential equation.

Question 4.
Form the differential equation not containing the arbitrary constant(s) and satisfied by the following equations :
(i) y = ax³, a is arbitrary constant
(ii) x² – y² = a², a is arbitrary constant
(iii) y = c sin^-1 x, c is arbitrary constant
(iv) y² = 4ax, a is arbitrary constant.
Solution:
(i) Given eqn. of curve be, y = ax³ ……………(1)
Diff. eqn. (1) w.r.t. x; we have
\(\frac{d y}{d x}\) = 3ax² ……………..(2)
on dividing eqn. (2) by eqn. (1); we have
\(\frac{1}{y} \frac{d y}{d x}=\frac{3 a x^2}{a x^3}=\frac{3}{x}\)
=> x \(\frac{d y}{d x}\) = 3y be the required differential eqn.

(ii) Given eqn. of curve be x² – y² = a² ……………(1)
Diff. eqn. (1) w.r.t. x; we have dy
\(\frac{d y}{d x}=\frac{c}{\sqrt{1-x^2}}\) ………….(2)
=> x – y \(\frac{d y}{d x}\) = 0 be the reqd. differential equation.

(iii) Given y = c sin^-1 x, ………………..(1)
where c be arbitrary constant
Differentiating w.r.t. x ; we have
\(\frac{d y}{d x}=\frac{c}{\sqrt{1-x^2}}\) ………………(2)
From (1) ;
c = \(\frac{y}{\sin ^{-1} x}\)
from (2) ;
\(\frac{d y}{d x}=\frac{y}{\sqrt{1-x^2} \sin ^{-1} x}\)
be the required diff. eqn.

(iv) Given eqn. of curve be, y² = 4ax …………..(1)
diff. eqn. (1) w.r.t. x; we have dy
2y \(\frac{d y}{d x}\) = 4a
From (1);
y² = 2xy \(\frac{d y}{d x}\)
=> 2x \(\frac{d y}{d x}\) = y be the required differential equation.

Find the differential equations by eliminating the arbitrary constants and satisfied by the following (5 to 11) equations :

Question 5.
(i) y = e^mx, m is arbitrary constant
(ii) y = k e^{tan-1 x}, k is arbitrary constant
(iii) y = tan^-1 x + ce^{– tan-1} x, c is arbitrary constant.
Solution:
(i) Given eqn. of curven be, y = e^mx ………………(1)
Diff. (1) w.r.t. x; we have dy dx dy
=> \(\frac{d y}{d x}\) = e^mx . m
=> \(\frac{d y}{d x}\) = my …………….(2) [Using (!)]
also from (1) ;
log y = mx
=> m = \(\frac{1}{x}\) log y
∴ From (2);
\(\frac{d y}{d x}=\frac{y \log y}{x}\)
=> x = y \(\frac{d y}{d x}\) the required differential equation.

(ii) Given y = k e^{tan-1 x} ……………….(1)
where k is arbitrary constant.
On differentiating eqn. (1) w.r.t. x; we have
\(\frac{d y}{d x}\) = k e^{tan-1 x} \(\frac{1}{1+x^2}\)
=> (1 + x²) be the required diff. eqn.

(iii) Given y = tan^-1 x + c e^{– tan-1 x}
where c be arbitrary constant
Diff. eqn. (1) w.r.t. x ; we have dy_
\(\frac{d y}{d x}=\frac{1}{1+x^2}+c e^{-\tan ^{-1} x}\left(-\frac{1}{1+x^2}\right)\)
=> (1 + x²) \(\frac{d y}{d x}\) = 1 – ce^{– tan-1 x}
=> (1 + x²) \(\frac{d y}{d x}\) + y = 1 – (y – tan^-1 x) [using eqn. (1)]
=> (1 + x²) \(\frac{d y}{d x}\) + y = 1 + tan^-1 x be the required diff. eqn.

Question 6.
(i) x² + (y – b)² = 1, b is arbitrary constant
(ii) (x – a)² – y² = 1, a is arbitrary constant
(iii) x² + y² = ax³, a is arbitrary constant.
Solution:
(i) Given eqn. of curve be
x² + (y – b)² = 1 …………………(1)
Diff. eqn. (1) w.r.t. x; we have
2x + 2 (y – b) \(\frac{d y}{d x}\) = 0
=> (y – b) = \(\frac{-x}{\frac{d y}{d x}}\)
From (1);
x² + \(\frac{x^2}{\left(\frac{d y}{d x}\right)^2}\) = 1
=> x + \(\left[1+\left(\frac{d y}{d x}\right)^2\right]=\left(\frac{d y}{d x}\right)^2\) be the required differential equation.

(ii) Given eqn. of curve be,
(x – a)² – y² = 1 …………………..(1)
Diff. eqn. (1) w.r.t. x; we get
2 (x – a) – 2y \(\frac{d y}{d x}\) = 0
=> x – a = y \(\frac{d y}{d x}\)
From (1);
(y \(\frac{d y}{d x}\))² – y² = 1
y² (\(\frac{d y}{d x}\))² – y² = 1 be the required differential equation.

(iii) Given eqn. of curve be, x² + y² = ax³ ………………..(1)
Diff. eqn. (1) w.r.t. x; we have
2x + 2y \(\frac{d y}{d x}\) = 3ax² ……………………(2)
Using eqn. (1) in eqn. (2); we have
2x + 2y \(\frac{d y}{d x}\) = 3 \(\left[\frac{x^2+y^2}{x}\right]\)
⇒ 2x² + 2xy \(\frac{d y}{d x}\) = 3x² + 3y²
⇒ x² + 3y² = 2xy \(\frac{d y}{d x}\)
be the required differential equation.

Question 7.
\(\frac{x}{a}+\frac{y}{b}\) = 1, a, b are arbitrary constants. (NCERT)
Solution:
Given \(\frac{x}{a}+\frac{y}{b}\) = 1 ……………….(1)
where a and b are arbitrary constants
Diff. eqn. (1) w.r.t. x ; we get
\(\frac{1}{a}+\frac{1}{b} \frac{d y}{d x}\) = 0
⇒ \(\frac{d y}{d x}=-\frac{b}{a}\)
On differentiating both sides w.r.t. x ; we have
\(\frac{d^2 y}{d x^2}\) = 0 be the required diff. eqn.

Question 8.
(i) y² = 4a (x – b), a, b are arbitrary constants.
(ii) (y – b)² = 4 (x – a), a, b are arbitrary constants.
(iii) y² = a (b² – x²), a, bare arbitrary constants. (NCERT)
Solution:
(i) Given eqn. of curve be
y² = 4a (x – b) ……………(1)
Diff. eqn. (1) w.r.t. x ; we have
2y \(\frac{d y}{d x}\) = 4a
again diff. eqn. (2) w.r.t. x ; we have
2 \(\left[y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2\right]\) = 0
⇒ y \(\frac{d^2 y}{d x^2}\) + (\(\frac{d y}{d x}\))² = 0 be the required differential equation.

(ii) Given (y – b)² = 4 (x – a)
Diff. (1) w.r.t. x ; we get
2 (y – b) y₁ = 4
again diff. both sides w.r.t. x, we get
2 [(y – b)y₂ + y₁²] = 0
Eliminating b from (2) and (3); we have
\(\frac{4}{y_1}\) . y₂ + 2 y₁² = 0
⇒ 2y₂ + y₁³ = 0 is the required differential equation.

(iiii) Given y² = a (b² – x)
Diff. eqn. (1) w.r.t. x ; we get
2yy₁ = – 2ax
⇒ \(\frac{y y_1}{x}\) = – a
diff. eqn. (2) w.r.t. x; we get
\(\frac{x\left(y y_2+y_1^2\right)-y y_1}{x^2}\) = 0
⇒ x (yy² + y¹²) – yy¹ = 0 be the reqd. diff. eqn.

Question 9.
(i) y = a cos (x + b), a, b are arbitrary constants.
(ii) y = A cos 2x + B sin 2x, A, B are arbitrary constants.
Solution:
(i) Given y = a cos (x + b) ………………….(1)
DiflF. (1) w.r.t. x ; we have
y’ = – a sin (x + b) …………………(2)
Diff. eqn. (2) w.r.t. x, we get
y” = – a cos (x + b) = -y [using (1)]
=> y” + y = 0 is the required diff. eqn.

(ii) Given y = a cos 2x + b sin 2x
where A, B are arbitrary constants.
On differentiating eqn. (1) w.r.t. x ; we have
\(\frac{d y}{d x}\) = – 2A sin 2x + 2B cos 2x ax
again differentiating w.r.t. x ; we have
\(\frac{d^2 y}{d x^2}\) = – 4A cos 2x – 4B sin 2x
= – 4 (A cos 2x + B sin 2x)
=> \(\frac{d^2 y}{d x^2}\) = – 4y [using eqn. (1)]
i.e. \(\frac{d^2 y}{d x^2}\) + 4y = 0 be the required diff. eqn.

Question 10.
(i) y = A e^2x + B e^{– 2x}, A, B are arbitrary constants. (NCERT Exemplar)
(ii) y = a e^3x + b e^{– 2x}, a, b are arbitrary constants. (NCERT)
Solution:
(i) Given eqn. of curve be y = A e^2x + B e^{– 2x} ……………….(1)
Diff. (1) w.r.t. x; we get
\(\frac{d y}{d x}\) = 2Ae^2x – 2Be^-2x ………………(2)
Again diff. (2) w.r.t. x, we have d2 v
\(\frac{d^2 y}{d x^2}\) = 4A e^2x + 4B e^{– 2x} = 4y [Using eqn. (1)]
be the required differential equation.

(ii) Given y = A e^3x + B e^{– 2x} ………….(1)
Diff. (1) w.r.t. x ; we get
y’ = 3A e^3x – 2 B e^{– 2x} ……………….(2)
eqn. (2) – 3 × eqn. (1) ; we have
y’ – 3y = – 5B e^{– 2x} ……………(3)
Diff. (2) w.r.t. x, we get
y” – 3y’ = + 10B e^{– 2x} ………………….(4)
eqn. (4) + 2 × eqn. (3); we have
y” – 3 y’ + 2 (y’ – 3y) = 0
=> y” – y’ – 6y = 0 is the required diff. eqn.

Question 11.
(i) y = (sin^-1 x)² + A cos^-1 x + B where A, B are arbitrary constants.
(ii) y = e* (A cos x + B sin x), A, B are arbitrary constants. (NCERT)
Solution:
(i) Given eqn. of curve be
y = (sin^-1 x)² + A(cos^-1 x) + B
where A and B are arbitrary constants
Diff. eqn. (1) w.r.t. x; we have

(ii) Given y = e^x (A cos x + B sin x) ……………..(1)
Diff. (1) w.r.t. x ; we get
=> y’ = e^x (A cos x + B sin x) + e^x (- A sin x + B cos x)
=> y’ = y + e^x (- A sin x + B cos x) ……………….(2)
Diff. eqn. (2) w.r.t. x; we have
y” = y’ + e^x (- A sin x + B cos x) + e^x (- A cos x – B sin x)
y” = y’ + y’ – y – y
Thus, y” – 2y’ + 2y = 0
be the required differential equation.

Question 12.
Find the differential equation of the family of curves (x – h)² + (y – k)² = r² where h, k are arbitrary constants.
Solution:
Given eqn. of curve be
(x – h)² + (y – k)² = r² ………………(1)
where h and k are parameters
diff. eqn. (1) w.r.t. x; we have
2(x – h) + 2 (y – k) = 0
⇒ (x – h) + (y – k) = 0
Again diff. eqn. (2) w.r.t. x; we have
1 + (y – k) \(\frac{d^2 y}{d x^2}\) + (\(\frac{d y}{d x}\))² = 0
∴From (3); we have
(y – k) = – \(\frac{1+p^2}{y_2}\) ;
where p = \(\frac{d y}{d x}\) & y₂ = \(\frac{d^2 y}{d x^2}\)
From (2), we have
(x – h) = – (y – k) \(\frac{d y}{d x}=\frac{\left(1+p^2\right) p}{y_2}\)
putting the values of (x – h) and (y – k) in eqn. (1); we have
\(\frac{\left(1+p^2\right)^2 p^2}{y_2{ }^2}+\frac{\left(1+p^2\right)^2}{y_2{ }^2}\) = r²
⇒ (1 + p²)³ = r²y₂² be the reqd. eqn.

Question 13.
(i) concentric circles with centre at (1, 2).
(ii) all circles which touch they-axis at origin.
(iii) circles in the second quadrant and touching the coordinate axes.
(iv) all parabolas having their vertices at origin and foci on y-axis.
(v) all ellipses whose centres are at origin and foci on y-axis. (NCERT)
(vii) all hyperbola having foci on x-axis and centres at the origin. (NCERT)
(vii) all non-vertical lines in a plane. (NCERT Exemplar)
Solution:
(i) eqn. of all concentric circles with centre at (1, 2) and radius r is given by
(x – 1)² + (y – 2)² = r² …………………..(1)
where r be arbitrary constant.
Duff. eqn. (1) w.r.t. x ; we have
2 (x – 1) + 2(y – 2) \(\frac{d y}{d x}\) = 0
⇒ (x – 1) + (y – 2) y₁ = 0 be the required differential eqn.

(ii) We know that eqn. of family of circles touching y-axis at origin is given by
(x – r)² + y² = r² ………………….(1)
where r be the arbitrary constant.

Diff. eqn. (1) w.r.t. x ; we have
2 (x – r) + 2yy’ = 0
⇒ x – r = – yy’
and r = x + yy’
∴ From eqn. (1) ; we have
(yy’)² + y² = (x + yy’)²
⇒ (yy’)² + y² = x² + (yy’)² + 2xyy’
⇒ x² – y² + 2xyy’ = 0 is the required differential eqn.

(iii) We know that. eqn. of family of circles touching coordinate axes lies in the 2nd quadrant is given by

(x + r)² + (y – r)² = r² …………(1)
where r be thc radius of any number of family of circles and is the only one arbitrary constant.
(1) can be written as
x² + y² + 2rx – 2ry + r² = 0 ……………(2)
Diff. (2) w.r.t. x; we have
2x + 2yy’ + 2r – 2ry’ = 0
⇒ \(\frac{x+y y^{\prime}}{-1+y^{\prime}}\) = + r
Putting in eqn. (1) ; we have
\(\left[x+\frac{x+y y^{\prime}}{y^{\prime}-1}\right]^2+\left[y-\frac{x+y y^{\prime}}{y^{\prime}-1}\right]^2=\left[\frac{x+y y^{\prime}}{y^{\prime}-1}\right]^2\)
⇒ \(\left[\frac{x y^{\prime}+y y^{\prime}}{y^{\prime}-1}\right]^2+\left[\frac{-x-y}{y^{\prime}-1}\right]^2=\left[\frac{x+y y^{\prime}}{y^{\prime}-1}\right]^2\)
⇒ y’² (x + y)² + (x + y)² = (x + yy’)²
⇒ (x + y)² (1 + y’)² = (x + yy’)² be the required diff. eqn.

(iv) equation of all parabolas having their vertices
at origin and foci on y-axis ¡s given by
x² = 4ay …………………..(1)
where a be arbitrary constant
Diff. eqn. (1) w.r.t. x ; we have
2x = 4a \(\frac{d y}{d x}\) ……………….(2)
On dividing (2) by (1) ; we have
\(\frac{2}{x}=\frac{\frac{d y}{d x}}{y}\)
⇒ x \(\frac{d y}{d x}\) = 2y be the required differential eqn.

(v) We know that eqn. of such family of ellipses is given by
\(\frac{x^2}{b^2}+\frac{y^2}{a^2}\) = 1 …………………(1) (a > b > 0)
where a and b are two arbitrary constants
Diff. (1) w.r.t, x ; we get
\(\frac{2 x}{b^2}+\frac{2 y y^{\prime}}{a^2}\) = 0
⇒ \(\frac{y y^{\prime}}{x}=-\frac{a^2}{b^2}\) ………………(2)
Diff. eqn. (2) w.r.t. x; we get
\(\frac{x\left(y y^{\prime \prime}+y^{\prime 2}\right)-y y^{\prime}}{x^2}\) = 0
xyy” + xy’² = yy’ is the required diff. eqn.

(vii) We know that eqn. of such family of hyperbolas is given by
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}\) = 1 ……………..(1) (a > 0, b > 0)
Diff. (1) w.r.t, x; we have
\(\frac{2 x}{a^2}-\frac{2 y y^{\prime}}{b^2}\) = 0
⇒ \(\frac{y y^{\prime}}{x}=\frac{b^2}{a^2}\) ………………….(2)
Diff. (2) w.r.t. x ; we have
\(\frac{x\left(y y^{\prime \prime}+y^{\prime 2}\right)-y y^{\prime}}{x^2}\) = 0
i.e. xyy” + y’² = yy’ is the required differential eqn.

(viii) We know that eqn. of any line in plane be given by
ax + by = 1
Now eqn. of any line || to y-axis be
x = constant
∴ b = 0, a ≠ 0
Thus eqn. of all non vertical lines in a plane is given by
ax + by = 1, b ≠ 0, a ∈ R
On differentiating w.r.t. x ; we have
a + b \(\frac{d y}{d x}\) = 0
again differentiating w.r,t. x ; we have
b \(\frac{d^2 y}{d x^2}\) = 0
⇒ \(\frac{d^2 y}{d x^2}\) = 0 [∵ b ≠ 0]
which is the required diff. eqn.

Question 13 (old).
(iv) all parabols having their vertices at origin and axis along positive direction of y-axis.
Solution:
Let (0, a) be the focus of any number of family of parabolas such.
∴ eqn. of such family of parabolas is given by
x² = 4ay ………………….(1)
Differential eqn. (1) w.r.t. x; we have
2x = 4ay’
⇒ y’ = \(\frac{x}{2a}\)
putting in eqn. (1); we get
x² = 4y \(\left(\frac{x}{2 y^{\prime}}\right)\)
⇒ xy’ = 2y is the required diff. eqn.

Question 14.
Form the differential equation of the family of circles having centres on y-axis and
(i) radius 3 units (NCERT)
(ii) passing through origin.
Solution:
(i) We know that eqn. of family of circle having centre on y-axis
i.e. (0, k) and radius 3 is given by
x² + (y – k)² = 3² = 9 …………..(1)
where k be arbitrary constant
Diff. (1) w.r.t. x ; we get
2x + 2 (y – k) y’ = 0
⇒ x + (y – k) y’ = 0
⇒ (y – k) = – \(\frac{x}{y^{\prime}}\)
∴ From (1);
x² + (- \(\frac{x}{y^{\prime}}\))² = 9
⇒ x² y’² = 9 is the required diff. eqn.

(ii) We know that, eqn. of circle passing through origin and centre on y-axis is
x² + y² + 2fy = 0 …………….(1)
f be any arbitrary constant.
Diff. (1) w.r.t. x ; we have
2x + 2yy’ + 2fy’ = 0
⇒ \(\frac{x+y y^{\prime}}{y^{\prime}}\) = – f
∴ From (1) ; we have
x² + y² + 2y \(\left[\frac{-x-y y^{\prime}}{y^{\prime}}\right]\) = 0
⇒ (x² + y²) y’ – 2xy – 2y²y’ = 0
⇒ (x² – y²) y’ = 2xy is the required differential equation.

Question 15.
Form the differential equation of simple harmonic motion given by x = A cos (nt + α), where n is fixed and A, α are parameters.
Solution:
Given x = A cos (nt + α) ………………(1)
where n is fixed. A, α are parameters
Diff. eqn. (1) w.r.t. t ; we have
\(\frac{d x}{d t}\) = – A sin (nt + α) . n
again differentiating w.r.t. t, we have
\(\frac{d^2 x}{d t^2}\) = – A cos (nt + α) . n² = – n²x [using eqn. (I)]
⇒ \(\frac{d^2 x}{d t^2}\) + n²x = 0, be the required diff. eqn.

Parents can use ML Aggarwal Maths for Class 12 Solutions to provide additional support to their children.

ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.2

Very short answer type questions (1 to 7) :

Question 1.
What is the number of arbitrary constants in the general solution of a differential equation of order 4 ? (NCERT)
Solution:
Since the general solution of the differential equation of order n contains n independent arbitrary constants.
Thus, the general solution of given diff. eqn. of order 4 contains n arbitrary constants.

Question 2.
What is the number of arbitrary constants in a particular solution of a differential equation of order 3 ? (NCERT)
Solution:
Since the particular solution of the diff. eqn. is obtained from general solution by giving particular values to all the arbitrary constants.
Hence the particular solution of a diff. eqn. of order 3 contains no arbitrary constants.

Question 3.
Verify that y = e^{– 3x} is a solution of the differential equation \(\frac{d y}{d x}\) + 3y = 0.
Solution:
Given y = e^{– 3x} ………………(1)
Diff. both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = – 3 e^{– 3x} = – 3y
⇒ \(\frac{d y}{d x}\) + 3y = 0, which is the given differential equaiton.
Hence y = e^{– 3x} be the soln. of given diff. eqn.

Question 3 (old).
What is the number of arbitrary constants in the general solution of the differential equation (1 + x²) (\(\frac{d^2 y}{d x^2}\))³ + y (\(\frac{d y}{d x}\))⁴ = 7 √x.
Solution:
Given diff. eqn. be,
(1 + x²) (\(\frac{d^2 y}{d x^2}\))³ + y (\(\frac{d y}{d x}\))⁴ = 7 √x
Since the order of given diff. eqn. be 2 as the highest order derivative existing in given diff. eqn. be \(\frac{d^2 y}{d x^2}\)
and its order be 2.
Thus the general solution of given diff. eqn. of order 2 contains 2 arbitrary constants.

Question 4.
In each of the following, show that the given function is a solution of the corresponding differential equation :
(i) y = cos x + C ; \(\frac{d y}{d x}\) + sin x = 0 (NCERT)
(ii) y = x² + 2x + C ; y’ – 2x – 2 = 0 (NCERT)
(iii) y = Ae^{– x} ; y’ + y = 0
(iv) y = Ax : xy’ = y (NCERT)
(v) y = e + 1 : y” – y’ = 0 (NCERT)
Solution:
(i) Given y = cos x + C …………..(1)
⇒ y’ = – sin x
⇒ y’ + sin x = 0 [Hence verified]

(ii) Given, y = x2 + 2x + c ………….(1)
Diff. eqn. (1) both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = 2x + 2
⇒ y’ – 2x – 2 = 0
which is the given diff. eqn.
Thus, y = x² + 2x + c be the solution of given differential eqn.

(iii) Given y = Ae^-x …………….(1)
Diff. eqn. (1) both sides w.r.t. x, we have
\(\frac{d y}{d x}\) = – Ae^-x = – y [using eqn. (1)]
⇒ y’ + y = 0, which is given diff. eqn.
Hence, y = Ae^-x is the solution of given differential eqn.

(iv) Given y = Ax ……………(1)
Diff. eqn. (1) both sides w.r.t. x, we have
y’ = A
⇒ y’ = \(\frac{y}{x}\) [using eqn. (1)]
⇒ xy’ – y = 0, which is given differential eqn.
Thus, y = Ax be the solution of given diff. eqn.

(v) Given y = e^x + 1
∴ y’ = e^x
⇒ y” = e^x
⇒ y” = y’
⇒ y” – y’ = 0.

Question 5.
Show that the function y = ax + 2a² is a solution of the differential equation 2(\(\frac{d y}{d x}\))² + x \(\frac{d y}{d x}\) – y = 0.
Solution:
Given y = ax + 2a²
L.H.S = 2(\(\frac{d y}{d x}\))² + x \(\frac{d y}{d x}\) – y
= 2a² + ax – (ax + 2a²) = 0
= R.H.S.

Question 7 (old).
Show that y = c₁ e^x + c₂ e^{– x} is the general solution of the differential equation \(\frac{d^2 y}{d x^2}\) – y = 0.
Solution:
Given y = c₁ e^x + c₂ e^{– x} …………….(1)
Differentiating (1) w.r.t. x, we have
\(\frac{d y}{d x}\) = c₁ e^x + c₂ e^{– x} …………….(2)
diff. eqn. (2) w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = c₁ e^x + c₂ e^{– x} = y [using eqn. (1)]
⇒ \(\frac{d^2 y}{d x^2}\) – y = 0, which is given diff. eqn.
Thus, y = c₁ e^x + c₂ e^{– x} be the solution of the given diff. eqn.

In each of the following (8 to 14), show that the given function (explicit or implicit) is a solution of the corresponding differential equation :

Question 8.
y = \(\sqrt{a^2-x^2}\), x ∈(- a, a) : x + y \(\frac{d y}{d x}\) = 0. (NCERT)
Solution:
Given y= \(\sqrt{a^2-x^2}\) ;
on squaring
⇒ y² = a² – x²
⇒ x² + y² = a²
Diff. both sides w.r.t. x ; we have
2x + 2y \(\frac{d y}{d x}\) = 0
which is the same as given differential equation
Thus y = \(\sqrt{a^2-x^2}\) be the solution of x + y \(\frac{d y}{d x}\) = 0.

Question 9.
y² = 4ax : y = x \(\frac{d y}{d x}\) + a \(\frac{d x}{d y}\)
Solution:
Given y² = 4ax …………..(1).
Diff. eqn. (1) w.r.t. x ; we have
2y \(\frac{d y}{d x}\) = 4a
⇒ \(\frac{d y}{d x}=\frac{4 a}{2 y}=\frac{2 a}{y}\)
Now R.H.S.
⇒ \(x \frac{d y}{d x}+a \frac{d y}{d x}=\frac{2 a x}{y}+\frac{a y}{2 a}\)
= \(\frac{y^2}{2 y}+\frac{y}{2}\) [∵ Using (1)]
= y
= L.H.S
THus y² = 4ax be the solution of x \(\frac{d y}{d x}\) + a \(\frac{d x}{d y}\) = 1.

Question 9 (old).
y = \(\frac{1}{4}\) (x ± A)² : y₁² = y.
Solution:
Given y = \(\frac{1}{4}\) (x ± A)²
Diff. both sides w.r.t. x ; we have
\(\frac{d y}{d x}\) = \(\frac{1}{2}\) (x ± A)
On squaring both sides, we have
(\(\frac{d y}{d x}\))² = \(\frac{1}{4}\) (x ± A)² = y
Thus y = \(\frac{1}{4}\) (x ± A)² be the solution of given differential eqn (\(\frac{d y}{d x}\))² = y.

Question 10.
x + y = tan^-1 y : y²y’ + y^-1 + 1 = 0
Solution:
Given, x + y = tan^-1 y ……………(1)
Diff. eqn. (1) both sides w.r.t. x, we have
1 + y’ = \(\frac{1}{1+y^2}\) y’
⇒ (1 + y²) (1 + y’) = y’
⇒ 1 + y² + (1 + y² – 1) = y’
⇒ 1 + y² + y²y’ = 0 which is given diff. eqn.
Thus, x + y = tan^-1 y is the solution of given diff. eqn.

Question 11.
y = a cos x + b sin x ; \(\frac{d^2 y}{d x^2}\) + y = 0 (NCERT)
Solution:
Given y = a cos x + b sin x ………………(1)
Diff. both sides of eqn. (1) w.r.t. x ; we have
\(\frac{d y}{d x}\) = – a sin x + b cos x
again diff. w.r.t. x ; we have
\(\frac{d^2 y}{d x^2}\) = – a cos x – b sin x = – y
⇒ \(\frac{d^2 y}{d x^2}\) + y = 0 ……………..(2)

Question 12.
y = x sin 3x : \(\frac{d^2 y}{d x^2}\) + 9y – 6 cos 3x = 0 (NCERT)
Solution:
Given y = x sin 3x ……………….(1)
Diff. eqn. (1) both sides w.r.t. x ; we have
\(\frac{d y}{d x}\) = sin 3x + 3 x cos 3x ……………..(2)
Diff. eqn. (2) w.r.t. x ; we have
\(\frac{d^2 y}{d x^2}\) = 3 cos 3x + 3 (- 3x sin 3x + cos 3x)
⇒ \(\frac{d^2 y}{d x^2}\) = – 9x sin 3x + 6 cos 3x
⇒ \(\frac{d^2 y}{d x^2}\) + 9y – 6 cos 3x = 0 [using eqn. (1)]
Thus, y = x sin 3x be the solution of diff. eqn.

Question 13.
Show that y = e^-x + ax + b is a solution of the differential equation e^x \(\frac{d^2 y}{d x^2}\) = 1.
Solution:
Given y = e^-x + ax + b ………………..(1)
Diff. (1) w.r.t. x; we have
\(\frac{d y}{d x}\) = – e^-x + a ………….(2)
again diff. w.r.t x; we have
\(\frac{d^2 y}{d x^2}\) = + e^-x
e^x \(\frac{d^2 y}{d x^2}\) which is same as given differential equation.
Thus y = e^-x + ax + b be the solution of e^x \(\frac{d^2 y}{d x^2}\) = 1.

Question 13 (old).
y = e^{– 3x} : \(\frac{d^2 y}{d x^2}\) + \(\frac{d y}{d x}\) – 6y = 0
Solution:
Given y = e^{– 3x} ……………………(1)
\(\frac{d y}{d x}\) = – 3 e^{– 3x} ……………….(2)
Diff. eqn. (2) w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = 9 e^{– 3x} ………………(3)
Now, \(\frac{d^2 y}{d x^2}\) + \(\frac{d y}{d x}\) – 6y
= 9e^{– 3x} – 3e^{– 3x} – 6e^{– 3x} = 0
[using eqn. (1), (2) and (3)]
Hence y = e^{– 3x} be the solution of given differential eqn.

Question 14.
Show that y = a cos (log x) + b sin (log x) is a solution of the differential equation x² \(\frac{d^2 y}{d x^2}\) + x \(\frac{d y}{d x}\) + y = 0.
Solution:
Given y = a cos (log x) + b sin (log x) …………………(1)
Differentiating eqn. (1) w.r.t. x ; we have
\(\frac{d y}{d x}\) = – a sin (log x) . \(\frac{1}{x}\) + b cos (log x) . \(\frac{1}{x}\)
⇒ x \(\frac{d y}{d x}\) = – a sin (log x) + b cos (log x) …………………(2)
Differentiating eqn. (2) w.r.t. x ; we have
x \(\frac{d^2 y}{d x^2}\) + \(\frac{d y}{d x}\) . 1 = – a cos (log x) . \(\frac{1}{x}\) – b sin (log x) . \(\frac{1}{x2}\)
⇒ x² \(\frac{d^2 y}{d x^2}\) + x \(\frac{d y}{d x}\) = – y [using eqn. (1)]
⇒ x² \(\frac{d^2 y}{d x^2}\) + x \(\frac{d y}{d x}\) + y = 0, which is given diff. eqn.
Thus, y = a cos (log x) + b sin (log x) be the soln. of given diff. eqn.

Question 15.
Show that the differential equation of which y = 2 (x² – 1) + ce^{– x2} is a solution \(\frac{d y}{d x}\) + 2xy = 4x³.
Solution:
Given y = 2 (x² – 1) + ce^{– x2} ……………..(1)
Diff. both sides w.r.t. x; we have
∴ \(\frac{d y}{d x}\) = 4x + cce^{– x2} (- 2x)
\(\frac{d y}{d x}\) = 4x – 2x [y – 2x² + 2]
\(\frac{d y}{d x}\) = 4x – 2xy + 4x³ – 4x
\(\frac{d y}{d x}\) + 2xy = 4x³
which is same as given differential eqn.
Thus eqn. (1) be the solution of \(\frac{d y}{d x}\) + 2xy = 4x³..

Question 16.
Show that y² = 4a (x + a) is a solution of the differential equation y (1 – y₁²) = 2xy₁.
Solution:
Given y² = 4a (x + a) ……………….(1)
Diff. both sides of eqn (1) w.r.t. x; we have
2y \(\frac{d y}{d x}\) = 4a
⇒ \(\frac{d y}{d x}=\frac{2 a}{y}\) ……………..(2)

Thus y² = 4 a (x + a) be the solution of y \(\left\{1-\left(\frac{d y}{d x}\right)^2\right\}=2 x \frac{d y}{d x}\).

Question 17.
Show that y = sin (sin x) is a solution of the differential equation \(\frac{d^2 y}{d x^2}\) + tan x \(\frac{d y}{d x}\) + y cos² x = 0.
Solution:
Given, y = sin (sin x) …………………….(1)
Diff. eqn. (1) both sides w.r.t. x ; we have
\(\frac{d y}{d x}\) = cos (sin x) cos x ………………….(2)
Diff. eqn. (2) both sides w.r.t. x, we have
\(\frac{d^2 y}{d x^2}\) = cos (sin x) (- sin x) + cos² x (- sin (sin x)) …………….(3)
From (2);
\(\frac{1}{\cos x} \frac{d y}{d x}\) = cos (sin x)
∴ from (3); we have
\(\frac{d^2 y}{d x^2}\) = – tan x \(\frac{d y}{d x}\) – y cos² x (using (1)]
⇒ \(\frac{d^2 y}{d x^2}\) + tan x \(\frac{d y}{d x}\) + y cos² x = 0
which is given diff. eqn.
Thus, y = sin (sin x) be the solution of given diff. eqn.

Question 18.
Show that y = e^{m sin-1 x} is a solution of the differential equation (1 – x²) y₂ – xy₁ – m²y = 0.
Solution:
Given y = e^{m sin-1 x} ……………………..(1)
Differentiating eqn. (1) w.r.t. x; we have
\(\frac{d y}{d x}\) = e^{m sin-1 x} \(\frac{m}{\sqrt{1-x^2}}\)
⇒ \(\sqrt{1-x^2} \frac{d y}{d x}\) = my ………………..(2) [using eqn. (1)]
Differentiating eqn. (2) w.r.t. x ; we have
\(\sqrt{1-x^2} \frac{d^2 y}{d x^2}+\frac{1}{2}\left(1-x^2\right)^{-\frac{1}{2}}(-2 x) \frac{d y}{d x}=m \frac{d y}{d x}\)
⇒ \(\sqrt{1-x^2} \frac{d^2 y}{d x^2}-\frac{x}{\sqrt{1-x^2}} \frac{d y}{d x}=m \sqrt{1-x^2} \frac{d y}{d x}\)
⇒ (1 – x²) y₂ – xy₁ = m²y [using eqn. (2)]
⇒ (1 – x²) y₂ – xy₁ – m²y = 0, which is given diff. eqn.
Thus, y = e^{m sin-1 x} be the soln. of given diff. eqn.