OP Malhotra Class 12 Maths Solutions Chapter 17 Differential Equations Ex 17(d)

The availability of ISC Class 12 OP Malhotra Solutions Chapter 17 Differential Equations Ex 17(d) encourages students to tackle difficult exercises.

S Chand Class 12 ICSE Maths Solutions Chapter 17 Differential Equations Ex 17(d)

Solve the following differential equations:

Question 1.
(i) \(\frac { dy }{ dx }\) = e^x+y
(ii) \(\frac { dy }{ dx }\) = x³ e^-2y
Solution:
(i) Given \(\frac { dy }{ dx }\) = e^x+y = e^x e^y
⇒ e^-y dy = e^x dx
[after variable separation]
On integrating ; we have
∫ e^-y dy = ∫ e^x dx + c
⇒ – e^-y = e^x + c ⇒ e^x + e^-y = c’
which is the required solution.

(ii) Given \(\frac { dy }{ dx }\) = x³ e^-2y

Question 2.
\(\frac{d y}{d x}=\frac{x y+y}{x y+x}\)
Solution:

Question 3.
y (1 – x²) dy = x(1 + y²) dx
Solution:
Given y (1 – x²) dy = x (1 + y²) dx

Question 4.
\(\frac{d y}{d x}+\frac{\cos x \sin y}{\cos y}\) = 0
Solution:
Given \(\frac{d y}{d x}+\frac{\cos x \sin y}{\cos y}\) = 0 ;
after variation separation, we have
\(\frac{\cos y}{\sin y}\)+ cos x dx = 0 ; On integrating
⇒ ∫ cot y dy + ∫ cos xdx = c
⇒ log sin y + sin x = c
⇒ log sin y = c – sin x
⇒ sin y = Ae^{-sin x}
which is the required solution.
[where A = e^c]

Question 5.
(i) (x + xy) dx + (x – xy²) dy = 0
(ii) (x² – yx²) dy + (y²+ xy²) dx = 0
(iii) x² (y + 1) dx + y² (x- 1) dy = 0
Solution:
(i) (y + xy) dx + (x – xy²) dy = 0
⇒ y (1 + x) dx + x (1 – y²) dy = 0
⇒ \(\left(\frac{1+x}{x}\right) d x+\left(\frac{1-y^2}{y}\right)\)dy = 0
On integrating ; we have
\(\int\left(\frac{1}{x}+1\right) d x+\int \frac{1}{y} d y-\int y d y\) = c
⇒ log |x| + x + log |y | – \(\frac { y² }{ 2 }\) = c
be the required solution

(ii) (x² – yx²) dy + (y² + xy²) dx = 0
⇒ x² (1 – y) dy + y² (1 + x) dx = 0
⇒ \(\frac{1-y}{y^2} d y+\frac{1+x}{x^2}\) = 0
On integrating ; we have
\(\int\left(\frac{1}{y^2}-\frac{1}{y}\right) d y+\int \frac{1}{x^2} d x+\int \frac{1}{x} d x\) = 0
⇒ – \(\frac{1}{y}-\log |y|-\frac{1}{x}+\log |x|\)
⇒ \(\log \frac{x}{y}=\frac{x+y}{x y}\) + c
be the required solution.

(iii) Given x² (y + 1) dx + y² (x – 1) dy = 0

be the required solution.

Question 6.
\(\sqrt{a+x} \frac{d y}{d x}\) = – xy
Solution:
Given \(\sqrt{a+x} \frac{d y}{d x}\) = – xy ; after variable separation, we have

which is the required solution.

Question 7.
\(\frac { dy }{ dx }\) = 1 – x + y – xy
Solution:
Given \(\frac { dy }{ dx }\) = 1 – x + y – xy
= 1 – x + y(1 – x)
⇒ \(\frac { dy }{ dx }\) = (1 – x)(1 + y)
After variable separation, we have \(\frac { dy }{ 1+y }\) = (1 – x) dx ; on integrating
log | 1 + y | = x – \(\frac { x² }{ 2 }\) + c which is the required solution.

Question 8.
(1 + x) (1 + y²) dx + (1 +.y) (1 + x²) dy = 0
Solution:
Given (1 + x) (1 + y²) dx + (1 + y) (1 + x²) dy = 0

Question 9.
see² x tany dx – sec²y tan x dy = 0
Solution:
Given sec² x tan y dx – sec² y tan x dy = 0
On dividing throughout by tan x tan y; we get
\(\frac{\sec ^2 x d x}{\tan x}-\frac{\sec ^2 y d y}{\tan y}\) = 0 ; on integrating
\(\int \frac{\sec ^2 x d x}{\tan x}-\int \frac{\sec ^2 y d y}{\tan y}\) = log c
⇒ log | tan x | – log | tan y | = log c
⇒ log \(\left|\frac{\tan x}{\tan y}\right|\) = log c
⇒ tan x = A tan y where A = ± c which is the required solution.

Question 10.
\(\frac{d y}{d x}=e^{x-y}+e^{2 \log x-y}\)
Solution:

Question 11.
(i) cos x cos y dy + sin x sin y dx = 0
(ii) (1 + cos x) dy = ( 1 – cos y) dx
Solution:
(i) Given cos x cos y dy + sin x sin y = 0
after variable separation, we have
\(\frac{\cos y d y}{\sin y}+\frac{\sin x d x}{\cos x}\) = 0; On integrating
= log |sin y| – log |cos x| = log c
⇒ log \(\left|\frac{\sin y}{\cos x}\right|\) = log c
⇒ sin y = ± c cos x
⇒ sin y = A cos x
which is the required solution.

(ii) Given (1 + cos x)dy = (1 – cos y) dx
after variable separation, we have

Question 12.
(1 + x²)dy + x\(\sqrt{1-y^2}\) dx = 0
Solution:

Question 13.
(1 – x²) dy + xy dx = xy² dx
Solution:

Question 14.
\(x \sqrt{1+y^2} d x+y \sqrt{1+x^2}\)dy = 0
Solution:
Given diff. eqn. be, \(x \sqrt{1+y^2} d x+y \sqrt{1+x^2}\)dy = 0
dividing throughout by \(\sqrt{1+x^2} \sqrt{1+y^2}\); we have

which is the required solution.

Question 15.
(e^x + 1)y dy = (y + 1)e^xdx
Solution:
Given (e^x + 1 )y dy = (y + 1)e^x dx; after variable separation, we have

which is the required solution.

Question 16.
(i) log\(\left(\frac{d y}{d x}\right)\) = ax + by
(ii) log\(\left(\frac{d y}{d x}\right)\) = 3x – 5y
Solution:

Question 17.
\(\frac { dy }{ dx }\) = cos³ x sin⁴ x + x\(\sqrt{2 x+1}\)
Solution:
Given \(\frac { dy }{ dx }\) = cos³ x sin⁴ x + x\(\sqrt{2 x+1}\) after variable separation, we have

Question 18.
\(\sqrt{1+x^2+y^2+x^2 y^2}+x y \frac{d y}{d x}\) = 0
Solution:
Given diff. eqn. be, \(\sqrt{1+x^2+y^2+x^2 y^2}+x y \frac{d y}{d x}\) = 0

Question 19.
\(\frac{d y}{d x}=\frac{e^x\left(\sin ^2 x+\sin 2 x\right)}{y(2 \log y+1)}\)
Solution:
Given, \(\frac{d y}{d x}=\frac{e^x\left(\sin ^2 x+\sin 2 x\right)}{y(2 \log y+1)}\) after variable separation, we have

Question 20.
\(\sqrt{1+x^2} d y+\sqrt{1+y^2} d x\) = 0
Solution:

Question 21.
\(\frac{d y}{d x}-x \sin ^2 x=\frac{1}{x \log x}\)
Solution:

Question 22.
Find the particular solution of the following differential equations :
(i) cos y dy + cos x sin y dx = 0, y(\(\frac { π }{ 2 }\)) = \(\frac { π }{ 2 }\)
(ii) (1 – x²)\(\frac { dy }{ dx }\) – xy = x; given y = 1 when x = 0.
(iii) (x² – yx²)dy + (y² + x²y²)dx = 0, given that y = 1, when x = 1.
(iv) \(\frac { dy }{ dx }\) = 1 + x² + y² + x²y², given that y = 1 when x = 0.
(v) (1 + e^2x) dy + (1 + y²)e^xdx = 0, given that y = 1 when x = 0.
(vi) \(\frac { dy }{ dx }\) = 1 + x + y + xy, given that y = 1 when x = 0.
(vii) \(\frac { dy }{ dx }\) = \(\frac{x(2 \log |x|+1)}{\sin y+y \cos y}\), given that y = \(\frac { π }{ 2 }\), when x = 1.
Solution:
(i) Given cos y dy + cos x sin y dx = 0 after variable separation, we have

(ii) Given (1 – x²)\(\frac { dy }{ dx }\) – xy = x ⇒ (1 – x²)\(\frac { dy }{ dx }\) = x (y + 1)

(iii) Given diff. eqn. be, (x² – yx²) dy + (y² + x²y²) dx = 0
⇒ x² (1 – y) dy + y² (1 + x²) dx = 0 ; after variable separation, we have

(iv) Given diff. eqn. be, \(\frac { dy }{ dx }\) = 1 + x² + y² + x²y² = (1 + x²)(1 + y²);
after variable separation; we have

(v) Given diff. eqn. be, (1 + e^2x) dy + (1 + y²)e^x dx = 0 ; after variable separation, we have

(vi) Given \(\frac { dy }{ dx }\) = 1 + x + y + xy = (1 + x) (1 + y) ⇒ \(\frac { dy }{ 1+y }\) = (1 + x)dx;

Question 23.
\(\frac { dy }{ dx }\) = (2x + 3y – 4)²
Solution:
Given \(\frac { dy }{ dx }\) = (2x + 3y – 4)²
put 2x + 3y – 4 = t
Diff. both sides w.r.t. x

which is the required solution.

Question 24.
(x + y + 1)\(\frac { dy }{ dx }\) = 1
Solution:
Given diff. eqn. be,

where c’ = c – 1

Question 25.
(x + y)² \(\frac { dy }{ dx }\) = 1
Solution:
Given diff. eqn. be,

which the required solution.

Question 26.
\(\frac { dy }{ dx }\) = cos(x + y)
Solution:

Question 27.
\(\frac { dy }{ dx }\) = tan²(x + y)
Solution:
Given diff. eqn. be,

where c’ = 4c
which is the required solution.

Question 28.
cos²(x – 2y) = 1 – 2\(\frac { dy }{ dx }\)
Solution:
Given differential eqn. be,

which is the required solution.

Question 29.
\(\frac { dy }{ dx }\) = \(\frac{x+y+1}{2 x+2 y+3}\)
Solution:
Given differential eqn. be,
\(\frac { dy }{ dx }\) = \(\frac{x+y+1}{2 x+2 y+3}\) … (1)
put x + y = t; diff. both sides w.r.t. x ; we get

which is the required solution.