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|
= elog |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
= $$e^{\int\left(\frac{1}{x}-\cot x\right) d x}$$
= elog 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 ey dx = (y3 + 2x ey) 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
= elog 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
= elog 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
= $$e^{\int \frac{y}{1-y^2} d y}$$
= $$e^{-\frac{1}{2} \int \frac{-2 y d y}{1-y^2}}$$
= $$e^{-\frac{1}{2} \log \left(1-y^2\right)}$$
= elog (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
= elog 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
= $$\int_e \frac{2}{x} d x=e^{2 \log x}$$
= elog 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 ex dy + (y ex + 2x) dx = 0. (NCERT)
Solution:
Given differential eqn. can be written as,
ex dy + (y ex + 2x) dx = 0
⇒ ex $$\frac{d y}{d x}$$ + y ex + 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
= ex
and solution is given by
y . e∫ P dx = ∫ (Q . e∫ P dx) dx + C
⇒ y . ex = ∫ – $$\frac{2 x}{e^x}$$ . ex dx + C
⇒ y ex = – x2 + C, which gives the required solution.

Question 6.
(i) $$\frac{d y}{d x}$$ + 2y = 6ex
(ii) $$\frac{d y}{d x}$$ + 3y = e– 2x (NCERT)
Solution:
(i) Given eqn. is $$\frac{d y}{d x}$$ + 2Y = 6ex …………….(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 = 6ex
∴ I.F = e∫ P dx
= e∫ 2 dx
= e2x
Multiply eqn. (1) by e2x; we get
e2x $$\frac{d y}{d x}$$ + 2 e2x y = 6 e3x
⇒ $$\frac{d}{d x}$$ (y e2x) = 6 e3x ;
On integrating we have
y e2x = ∫ 6 e3x dx + C
⇒ y e2x = $$\frac{6 e^{3 x}}{3}$$ + C
⇒ y = 2 ex + 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
= e3x
Multiply given eqn. by e3x, we get
e3x $$\frac{d y}{d x}$$ + 3 e3x y = e– 2x . e3x
⇒ $$\frac{d}{d x}$$ (e3x . y) = ex
On integrating ; we have
e3x . y = ∫ ex dx + C
⇒ e3x y = ex + 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
= elog 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) ex, 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) ex
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 = $$\frac{(x-1)}{x}$$ ex

(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 = ey
⇒ xey = ∫ (tan y + sec2 y) ey dy + C
= ∫ ey tan y dy + ∫ ey sec2 y dy + C
= tan y ey – ∫ sec2 y ey dy + ∫ ey sec2 y dy + C
⇒ x ey = tan y ey + 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
= $$e^{\int \frac{2}{x} d x}$$
= $$e^{2 \int \frac{1}{x} d x}$$
= e2 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. = $$e^{\int \frac{\cos x d x}{\sin x}}$$
= elog 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 = esin 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
= $$e^{\int \frac{\cos x d x}{1+\sin x}}$$
= elog (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 = esin x cos x

where P = cos x and Q = esin x cos x
Now I.F. = e∫ P dx
= e∫ cos x dx
= esin x
and solution is given by
y . e∫ P dx = ∫ Q . e∫ P dx dx + c
⇒ y . esin x = esin x cos x . esin x dx + c
= ∫ e2 sin x cos x dx + c
put sin x = t
⇒ cos x dx = dt
y esin x = ∫ e2 t dt + c
⇒ y esin x = $$\frac{e^{2 t}}{2}$$ + c
⇒ y esin x = $$\frac{1}{2}$$ e2 sin x + c
⇒ y = $$\frac{1}{2}$$ esin x + ce– sin x
which is the required solution curve.

Question 12.
(i) $$\frac{d y}{d x}$$ – tan x . y = esin 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 = esin x ; 0 < x < $$\frac{\pi}{2}$$
which is linear diff. eqn. in y
Here P = – tan x and Q = esin 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 = ∫ esin x cos x dx + C
put sin x = t
⇒ cos x dx = dt
y cos x = ∫ et dt + C
⇒ y cos x = esin x + C
⇒ y = sec x (esin 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
= ex
and solution is given by
y . e∫ P dx = ∫ Q . e∫ P dx + C
yex = ∫ ex (cos x – sin x) + C
= ∫ ex cos x dx – ∫ ex sin x dx + C
= cos x ex + ∫ sin x ex dx – ∫ ex sin x dx + C
⇒ yex = cos x ex + 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
= $$e^{\int \frac{1}{x} d x}$$
= elog 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
= $$e^{\int \frac{2}{x} d x}$$
= e2 log x
= elog 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
= elog |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
= $$e^{\int \frac{2 \cos x d x}{\sin x}}$$
= e2 log sin x
= elog 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
= ex2
and solution is given by
y . e∫ P dx = ∫ Q . e∫ P dx dx + C
∴ y . ex2 = ∫ x3 . ex2 dx + C
put x2 = t
⇒ 2x dx = dt
y ex2 = ∫ et . t $$\frac{d t}{2}$$ + C
yex2 = $$\frac{1}{2}$$ [t et – et] + C
yex2 = $$\frac{1}{2}$$ (x2 – 1) ex2 + C
y = $$\frac{1}{2}$$ (x2 – 1) + Cex2
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 ex
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
= $$e^{\int \frac{1 / x}{\log x} d x}$$
= elog (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
= $$e^{\int \frac{1}{y} d y}$$
= elog 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
= elog 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
= ey
and solution is given by
x . e∫ P dy = ∫ Q . e∫ P dy dy + C
⇒ xey = ∫ e– y log y . ey dy + C
⇒ xey = ∫ log y . 1 dy + C
⇒ xey = y log y – y + C
⇒ xey = 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
= elog 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 etan-1 y = ∫ et . t dt + C
= (t – 1) et + C
⇒ x etan-1 y = (tan-1 y – 1) etan-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 = 6ex …………….(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 = 6ex
∴ I.F = e∫ P dx
= e∫ 2 dx
= e2x
Multiply eqn. (1) by e2x; we get
e2x $$\frac{d y}{d x}$$ + 2 e2x y = 6 e3x
⇒ $$\frac{d}{d x}$$ (y e2x) = 6 e3x ;
On integrating we have
y e2x = ∫ 6 e3x dx + C
⇒ y e2x = $$\frac{6 e^{3 x}}{3}$$ + C
⇒ y = 2 ex + 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
= ex2
and solution is given by
y . e∫ P dx = ∫ Q . e∫ P dx dx + C
⇒ y . ex2 = ∫ x . ex2 dx + C
put x2 = t
⇒ 2x dx = dt
y . ex2 = $$\frac{1}{2}$$ ∫ et dt
⇒ y . ex2 = $$\frac{1}{2}$$ ex2 + 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
= etan x
and solution is given by
y . e∫ P dx = ∫ Q . e∫ P dx dx + C
⇒ y . etan x = ∫ tan x . sec2 x . etan x dx + C
put tan x = t
⇒ sec2 x dx = dt
y etan x = ∫ t et dt + C
⇒ y etan x = [t et – et] + C
⇒ y . etan x = (tan x – 1) etan 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 = ex, given that y = 1 when x = 0
(ii) y’ + y = ex, 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 = ex,
which is L.D.E. in y and is of the form
$$\frac{d y}{d x}$$ + Py = Q ;
where P = – 1 and Q = ex
∴ 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 = ∫ ex . 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) ex
which gives the required solution.

(ii) Given diff. eqn. be,
y’ + y = ex
which is L.D.E. and of the form
$$\frac{d y}{d x}$$ + Py = Q
where p = 1 and Q = ex
∴ I.F. = e∫ P dx
= e∫ dx dx
= ex
and its solution is given by
y . e∫ P dx =∫ Q . e∫ P dx dx + c
⇒ y . ex = ∫ ex . ex dx + c
= ∫ e2x dx + c
⇒ yex = $$\frac{e^{2 x}}{2}$$ + 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
= $$e^{\int \frac{1}{x} d x}$$
= elog 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
= elog 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 = et
⇒ dx = et dt
= $$\int \frac{t}{e^{2 t}}$$ et 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
= $$e^{\int \frac{2 x}{1+x^2} d x}$$
= elog (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
= e2x
and solution is given by
y . e∫ P dx = ∫ Q . e∫ P dx dx + c
⇒ y . e2x = ∫ e-2x sin x . e2x dx + c
⇒ ye2x = – 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
ye2x = 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
= $$e^{\int \frac{1}{x} d x}$$
= elog 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.