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

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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.