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

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

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

Very short answer type questions (1 to 6) :

Solve the following (1 to 20) differential equations:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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