Accessing Class 12 ISC Maths Solutions Chapter 9 Differential Equations Ex 9.6 can be a valuable tool for students seeking extra practice.

## ML Aggarwal Class 12 Maths Solutions Section A Chapter 9 Differential Equations Ex 9.6

Very Short answer type questions (1 to 2) :

Question 1.
Which of the following differential equations are homogeneous ?
(i) (x – y) $$\frac{d y}{d x}$$ = x + 2y
(ii) y – x $$\frac{d y}{d x}$$ = x + y $$\frac{d y}{d x}$$
Solution:
(i) Given (x – y) $$\frac{d y}{d x}$$ = x + 2y
⇒ $$\frac{d y}{d x}=\frac{x+2 y}{x-y}$$,
which is homogenenous diff. eqn.
Since $$\frac{d y}{d x}=\frac{x\left(1+\frac{2 y}{x}\right)}{x\left(1-\frac{y}{x}\right)}=\phi\left(\frac{y}{x}\right)$$
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
Thus from given eqn ; we have

(ii) Given differential eqn. be $$\frac{d y}{d x}=\frac{y-x}{y+x}$$
which is homogenenous diff. equation.
Since $$\frac{d y}{d x}=\frac{x\left(\frac{y}{x}-1\right)}{x\left(\frac{y}{x}-1\right)}=\phi\left(\frac{y}{x}\right)$$
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
∴ From eqn. (1) ; we have

⇒ $$\frac{1}{2}$$ log (v2 + 1) + tan-1 v + log x = $$\frac{1}{2}$$ log c
⇒ log (v2 + 1) + tan-1 v + log x = $$\frac{1}{2}$$ log c
⇒ log ($$\frac{y^2}{x^2}$$ + 1) x2 + 2 tan-1 ($$\frac{y}{x}$$) = A
⇒ log (x2 + y2) + 2 tan-1 ($$\frac{y}{x}$$) = A
which is the required solution.

Question 1 (old).
When a differential equation is called homogeneous ?
Solution:
A differential eqn. $$\frac{d y}{d x}$$ = f(x, y) is said to be homogeneous diff. eqn. iff(x, y) be a homogenenous function of degree 0.
i.e. a homogenenous diff. eqn. is of the form,
$$\frac{d y}{d x}$$ = f ($$\frac{y}{x}$$)
or $$\frac{d x}{d y}$$ = g ($$\frac{x}{y}$$)

Question 2.
(x2 + 3xy – 4y2) dx – x (x2 + 2y) dy = 0
Solution:
Given diff. eqn. can be written as,
$$\frac{d y}{d x}=\frac{x^2+3 x y-4 y^2}{x\left(x^2+2 y\right)}$$
= $$\frac{x^2\left[1+\frac{3 y}{x}-4\left(\frac{y}{x}\right)^2\right]}{x^2\left(x+\frac{2 y}{x}\right)}$$
i.e. $$\frac{d y}{d x}$$ is not a function of $$\frac{y}{x}$$.
Thus given diff. eeqn. is not homogenenous.

Question 3.
(i) x $$\frac{d y}{d x}$$ + y = x tan-1 $$\frac{y}{x}$$
(ii) x cos ($$\frac{y}{x}$$) $$\frac{d y}{d x}$$ = y cos ($$\frac{y}{x}$$) – 3x2
Solution:
(i) Given diff. eqn. be,
x $$\frac{d y}{d x}$$ + y = x tan-1 $$\frac{y}{x}$$
⇒ $$\frac{d y}{d x}$$ = – $$\frac{y}{x}$$ + tan-1 $$\frac{y}{x}$$
Thus $$\frac{d y}{d x}$$ = f($$\frac{y}{x}$$)
∴ given diff. eqn. be homogenenous diff. eqn.

(ii) Given diff. eqn. be,
x cos ($$\frac{y}{x}$$) $$\frac{d y}{d x}$$ = y cos ($$\frac{y}{x}$$) – 3x2
∴ $$\frac{d y}{d x}=\frac{y \cos \left(\frac{y}{x}\right)-3 x^2}{x \cos \left(\frac{y}{x}\right)}$$
= $$\frac{\frac{y}{x} \cos \left(\frac{y}{x}\right)-3 x}{\cos \left(\frac{y}{x}\right)}$$
which is not of the form, $$\frac{d y}{d x}$$ = f($$\frac{y}{x}$$)
Hence given diff. eqn. is not homogenenous.

Question 3 (old).
(ii) (x – y) dy – (x + y) dx = 0 (NCERT)
Solution:
Given $$\frac{d y}{d x}=\frac{x+y}{x-y}$$,
which is homogeneous diff. eqn.
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
Thus given eqn. becomes

Question 4.
(i) x $$\frac{d y}{d x}$$ – y = $$\sqrt{x^2+y^2}$$
(ii) (x2 + y2 sin $$\frac{x}{y}$$) $$\frac{d y}{d x}$$ = 2xy ex/y
Solution:
(i) Given, diff. eqn. be
x $$\frac{d y}{d x}$$ – y = $$\sqrt{x^2+y^2}$$
⇒ $$\frac{d y}{d x}=\frac{y}{x}+\sqrt{1+\left(\frac{y}{x}\right)^2}$$
= f ($$\frac{y}{x}$$)
Hence given diff. eqn. is homogeneous diff. eqn.

(ii) Given diff. eqn. be

Question 5.
(i) 2y e$$\frac{x}{y}$$ dx + (y – 2xe$$\frac{x}{y}$$) dy = 0
(ii) y dx + x log ($$\frac{y}{x}$$) dy – 2x dy = 0
Solution:
(i) Given, diff. eqn. be
2y e$$\frac{x}{y}$$ dx + (y – 2xe$$\frac{x}{y}$$) dy = 0
⇒ $$\frac{d x}{d y}=-\frac{\left(y-2 x e^{x / y}\right)}{2 y e^{x / y}}$$
= $$\frac{\frac{2 x}{y} e^{x / y}-1}{2 e^{x / y}}$$
which is of the form $$\frac{d x}{d y}$$ = f ($$\frac{x}{y}$$)
Hence given diff. eqn. be homogenenous.

(ii) Given differential eqn. can be written as
y dx = [2x – x log ($$\frac{y}{x}$$)] dy
$$\frac{d x}{d y}=\frac{2 x}{y}+\frac{x}{y} \log \frac{x}{y}$$
= f ($$\frac{x}{y}$$)
Thus, given differential eqn. be homogenenous.

Question 6.
(i) $$\frac{d y}{d x}=\frac{x+y}{x}$$ (NCERT)
(ii) (x – y) dy = (x + y) dx
Solution:
(i) Given diff. eqn. be,
$$\frac{d y}{d x}=\frac{x+y}{x}$$
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
∴ v + x $$\frac{d v}{d x}$$ = $$\frac{x+v x}{x}$$ = 1 + v
x $$\frac{d v}{d x}$$ = 1
⇒ dv = $$\frac{d x}{x}$$
On integrating ; we have
v = log x + C
y = x log |x| + Cx
which is the required solution.

(ii) Given $$\frac{d y}{d x}=\frac{x+y}{x-y}$$,
which is homogenenous diff. eqn.
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
Thus, given eqn becomes ;

Question 7.
(i) (x – y) y’ = x + 3y
(ii) (x + 2y) dx – (2x – y) dy = 0
Solution:
(i) Given, (x – y) y’ = x + 3y
⇒ y’ = $$\frac{x+3 y}{x-y}$$ …………….(1)
Also, y’ = $$\frac{d y}{d x}$$
= $$\frac{1+3 \frac{y}{x}}{1-\frac{y}{x}}$$
= f($$\frac{y}{x}$$ )
Thus eqn. (1) be homogenenous diff. eqn.
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
∴ from (1) ; we have

(ii) Given (x + 2y) dx – (2x – y) dy = 0
⇒ $$\frac{d y}{d x}=\frac{x+2 y}{2 x-y}$$,
which is homogenenous differential equation.
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$

Question 7 (old).
Prove that x2 – y2 = c (x2 + y2)2 is the general solution of the differential equation (x3 – 3xy2) dx = (y3 – 3x2y) dy where c is a parameter.
Solution:
Given differential eqn. can be written as
$$\frac{d y}{d x}=\frac{x^3-3 x y^2}{y^3-3 x^2 y}$$ ………………….(1)
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
∴ from (1) ;

Question 8.
(i) (y2 – 2xy) dx = (x2 – 2xy) dy
(ii) 2xy dx + (x2 + 2y2) dy = 0
Solution:
(i) Given, (y2 – 2xy) dx = (x2 – 2xy) dy
⇒ $$\frac{d y}{d x}=\frac{y^2-2 x y}{x^2-2 x y}$$ ……………..(1)
which is homogenenous differential equation
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
∴ From eqn. (1) ; we have

⇒ – log |v| – log |v – 1| = 3 log |x| + log c
log [v (v – 1) x3] = – log c
= log $$\frac{1}{c}$$
= log A
⇒ $$\frac{y}{x}\left(\frac{y}{x}-1\right)$$ x3 = A
⇒ $$\frac{y(y-x)}{x^2}$$ × x3 = A
⇒ xy2 – x2y = A be the required solution.

(ii) Given 2xy dx + (x2 + 2y2) dy = 0
⇒ $$\frac{d y}{d x}=-\frac{2 x y}{x^2+2 y^2}$$ ……………….(1)
which is homogenenous differential equation
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
∴ From eqn. (1) ; we have

Question 8 (old).
(ii) x $$\frac{d y}{d x}$$ = y (log y – log x + 1)
Solution:
Given $$\frac{d y}{d x}$$ = $$\frac{y}{x}$$ (log y – log x + 1)
⇒ $$\frac{d y}{d x}=\frac{y}{x}\left\{\log \frac{y}{x}+1\right\}$$ …………………….(1)
[∵ log a – log b = log $$\frac{a}{b}$$]
putting y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
Thus from eqn. (1) ; we have
v + x $$\frac{d v}{d x}$$ = v {log v + 1}
⇒ x $$\frac{d v}{d x}$$ = v log v
$$\frac{d v}{v \log v}=\frac{d x}{x}$$
integrating both sides, we have
$$\int \frac{d v}{v \log v}=\int \frac{d x}{x}$$
⇒ $$\int \frac{\frac{1}{v} d v}{\log v}=\int \frac{d v}{x}$$
⇒ log (log v) – log x = log c
⇒ log [log v] = log cx
⇒ log v = cx
⇒ log (y/x) = cx be the required solution.

Question 9.
(i) (x2 y2) dx + 2xy dy = 0 (NCERT)
(ii) x2 $$\frac{d y}{d x}$$ = x2 + xy + y2 (NCERT Exemplar)
Solution:
(i) Given, $$\frac{d y}{d x}=\frac{y^2-x^2}{2 x y}$$ ………….(1)

(ii) Given x2 $$\frac{d y}{d x}$$ = x2 + xy + y2
⇒ $$\frac{d y}{d x}=\frac{x^2+x y+y^2}{x^2}$$ ………………..(1)
which is homogenenous differential equation
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
∴ From eqn. (1) ; we have
v + x $$\frac{d v}{d x}$$ = $$\frac{x^2+v x^2+v^2 x^2}{x^2}$$
⇒ x $$\frac{d v}{d x}$$ = 1 + v + v2 – v
= 1 + v2
⇒ $$\frac{d v}{1+v^2}=\frac{d x}{x}$$ ;
on integrating
⇒ $$\int \frac{d v}{1+v^2}=\int \frac{d x}{x}$$
⇒ tan-1 v = log |x| + c
⇒ tan-1 $$\frac{y}{x}$$ = log |x| + c be the required solution.

Question 10.
(i) x $$\frac{d y}{d x}$$ + $$\frac{y^2}{x}$$ = y
(ii) 3x2 $$\frac{d y}{d x}$$ – 3xy = y2
Solution:
(i) Given, x $$\frac{d y}{d x}$$ + $$\frac{y^2}{x}$$ = y
⇒ x $$\frac{d y}{d x}$$ = y – $$\frac{y^2}{x}$$
= $$\frac{x y-y^2}{x}$$
⇒ $$\frac{d y}{d x}=\frac{x y-y^2}{x}$$ ………………(1)
Clearly $$\frac{d y}{d x}=\frac{y}{x}-\left(\frac{y}{x}\right)^2=f\left(\frac{y}{x}\right)$$
Thus, eqn. (1) be homogenenous differential equation.
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
∴ from (1) ;
v + x $$\frac{d v}{d x}$$ = $$\frac{v x^2-v^2 x^2}{x^2}$$
⇒ x $$\frac{d v}{d x}$$ = v – v2 – v = – v2
⇒ $$\frac{d v}{v^2}=-\frac{d x}{x}$$
on integrating ; we have
⇒ $$\int \frac{d v}{v^2}=-\int \frac{d x}{x}+\mathrm{C}$$
⇒ – $$\frac{1}{v}$$ = – log |x| + C
⇒ – $$\frac{x}{y}$$ = – log |x| + C
⇒ log |x| = C + $$\frac{x}{y}$$
⇒ |x| = ex/y ec
⇒ x = ± ec ex/y
= A ex/y which is the required solution.

(ii) Given diff. eqn. be written as ;
$$\frac{d y}{d x}=\frac{3 x y+y^2}{3 x^2}$$ …………..(1)
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
∴ from (1) ;

Question 11.
(i) $$\frac{d y}{d x}=\frac{y}{x}+\frac{\sqrt{x^2-y^2}}{x}$$, x > 0
(ii) $$x \frac{d y}{d x}-y=2 \sqrt{y^2-x^2}$$, x > 0
Solution:
(i) Given,
$$\frac{d y}{d x}=\frac{y}{x}+\frac{\sqrt{x^2-y^2}}{x}$$, x > 0 ……………….(1)
Also, $$\frac{d y}{d x}=\frac{y}{x}+\sqrt{1-\left(\frac{y}{x}\right)^2}=\phi\left(\frac{y}{x}\right)$$
Thus eqn. (1) be homogenenous diff. eqn.
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
∴ from (1) ;
v + x $$\frac{d v}{d x}$$ = v + $$\sqrt{1-v^2}$$
⇒ $$x \frac{d v}{d x}=\sqrt{1-v^2}$$
⇒ $$\frac{d v}{\sqrt{1-v^2}}=\frac{d x}{x}$$
On integrating ; we have
$$\int \frac{d v}{\sqrt{1-v^2}}=\int \frac{d x}{x}+\mathrm{A}$$
⇒ sin-1 v = log |x| + A
⇒ sin-1 ($$\frac{y}{x}$$) = log |x| + A
which is the required solution.

(ii) Given,
$$x \frac{d y}{d x}-y=2 \sqrt{y^2-x^2}$$, x > 0

Question 12.
$$\left(\frac{y}{x} \cos \frac{y}{x}\right)$$dx – $$\left(\frac{x}{y} \sin \frac{y}{x}+\cos \frac{y}{x}\right)$$ dy = 0
Solution:
Given,
$$\left(\frac{y}{x} \cos \frac{y}{x}\right)$$dx – $$\left(\frac{x}{y} \sin \frac{y}{x}+\cos \frac{y}{x}\right)$$ dy = 0

Question 13.
(i) (1 + ey/x) dy + ey/x (1 – $$\frac{y}{x}$$) dx = 0
(ii) 2yex/y dx – (y – 2xex/y) dy = 0
Solution:
(i) Given diff. eqn. can be written as ;
$$\frac{d y}{d x}=\frac{-e^{y / x}\left(1-\frac{y}{x}\right)}{1+e^{y / x}}$$ ………………(1)
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
∴ from (1) ; we have
v + x $$\frac{d v}{d x}$$ = $$\frac{-e^v(1-v)}{1+e^v}$$
⇒ x $$\frac{d v}{d x}$$ = $$\frac{-e^v(1-v)}{1+e^v}$$ – v
= $$\frac{-e^v+v e^v-v-v e^v}{1+e^v}$$
⇒ $$\frac{\left(1+e^v\right) d v}{v+e^v}=-\frac{d x}{x}$$
On integrating both sides ; we have
$$\int \frac{\left(1+e^v\right) d v}{v+e^v}=-\int \frac{d x}{x}$$ + log C
⇒ log |v + ev| = – log |x| + log C
[∵ ∫ $$\frac{f^{\prime}(x) d x}{f(d)}$$ = log |f(x)| + C]
⇒ log |(v + ev)x| = log C
⇒ log |y + x ey/x| = log C
⇒ y + x ey/x = A [∵ ± C = A]
which is the required solution.

(ii) Given diff. eqn. can be written as
$$\frac{d x}{d y}=\frac{\left(2 x e^{x / y}-y\right)}{2 y e^{x / y}}$$ …………….(1)
Here f (x, y) = $$\frac{2 x e^{x / y}-y}{2 y e^{x / y}}$$
∴ f (kx, ky) = $$\frac{2 k x e^{k x / k y}-k y}{2 k y e^{k x / k y}}$$ = k0 f (x, y)
∴ f (x, y) is a homogenenous function of degree 0.
put $$\frac{x}{y}$$ = v
⇒ x = yv
Diff. both sides w.r.t. y, we have
⇒ $$\frac{d x}{d y}=v+y \frac{d v}{d y}$$ ……………….(2)
Putting eqn. (2) in eqn. (1) ; we get

Question 14.
Solve the differential equation (x + y) dy + (x – y) dx = 0, given that y = 1 when x = 1. (NCERT)
Solution:
Given $$\frac{d y}{d x}=\frac{y-x}{x+y}$$ ……………..(1)
Here f (x, y) = $$\frac{y-x}{x+y}$$
∴ f(kx, ky) = $$\frac{k y-k x}{k x+k y}$$
= $$\frac{k(y-x)}{k(x+y)}$$
= k0 f (x, y)
Thus f (x, y) is homogenenous function of degree 0.
putting y = vx so that
$$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$ ……………..(2)
using eqn. (2) in eqn. (1) ; we get

Question 14 (old).
Find particular solutions of the following differential equations :
(i) 2xy + y2 – 2x2 = 0, given that y(1) = 2.
(ii) $$\frac{d y}{d x}-\frac{y}{x}$$ + cosec $$\left(\frac{y}{x}\right)$$ = 0 given that y = 0 when x = 1.
(iii) xey/x – y + x $$\frac{d y}{d x}$$ = 0, given that y(e) = 0.
Solution:
(i) Given 2xy + y2 – 2x2 = 0
⇒ $$\frac{d y}{d x}=\frac{2 x y+y^2}{2 x^2}$$ ……………(1)
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$ ……………(2)
Using eqn. (2) in eqn. (1) ; we have
v + x $$\frac{d v}{d x}$$ = $$\frac{2 v x^2+v^2 x^2}{2 x^2}$$
⇒ $$\frac{x d v}{d x}=\frac{2 v+v^2}{2}-v=\frac{v^2}{2}$$
On variable separation, we have
$$\frac{2}{v^2} d v=\frac{d x}{x}$$ ;
On integrating
– $$\frac{2}{v}$$ = log |x| + C
⇒ – $$\frac{2x}{y}$$ = log |x| + C …………….(1)
given y(1) = 2 i.e. when x = 1, y = 2
∴ from (1) ;
– 1 = C
∴ eqn. (1) becomes
– $$\frac{2x}{y}$$ = log |x| – 1
⇒ $$\frac{2x}{y}$$ = 1 – log |x|
⇒ y = $$\frac{2 x}{1-\log |x|}$$ ; x ≠ 0, 1
which is the required solution.

(ii) Given $$\frac{d y}{d x}-\frac{y}{x}$$ + cosec $$\left(\frac{y}{x}\right)$$ = 0
⇒ $$\frac{d y}{d x}$$ = $$\frac{y}{x}$$ – cosec $$\frac{y}{x}$$ ……………(1)
which is homogenenous diff. eqn.
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
∴ from (1) ; we have
v + x $$\frac{d v}{d x}$$ = v – cosec v
⇒ x $$\frac{d v}{d x}$$ = – cosec v
⇒ $$\frac{d v}{\ {cosec} v}=-\frac{d x}{x}$$ ;
on integrating ; we have
⇒ ∫ sin v dv = – ∫ $$\frac{d v}{x}$$
⇒ – cos v = – log |x| + c
⇒ – cos $$\frac{y}{x}$$ = – log |x| + c
Since y (1) = 0
i.e. when x = 1 ; y = 0
∴ from (2) ; we have
– 1 = 0 + c
⇒ c = – 1
∴ From (2) ; we have
1 – cos $$\frac{y}{x}$$ = – log |x|
⇒ cos ($$\frac{y}{x}$$) – 1 = log |x| which is the required solution.

(iii) Given xey/x – y + x $$\frac{d y}{d x}$$ = 0
⇒ $$\frac{d y}{d x}=\frac{y-x e^{y / x}}{x}$$
= $$\frac{y}{x}$$ – e$$\frac{y}{x}$$ ………………(1)
which is homogeneous diff. eqn.
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
∴ From (2) ;
– 1 + log |e| = c
from (1) ; we have
v + x $$\frac{d v}{d x}$$ = v – ev
⇒ x $$\frac{d v}{d x}$$ = – ev
⇒ e– v dv = – $$\frac{d x}{x}$$ ;
On integrating ; we have
∫ e– v dv = – ∫$$\frac{d x}{x}$$
⇒ – e– v + log |x| = c
⇒ – e– y/x + log |x| = c ……………..(2)
Since y (e) = 0
i.e. When x = e ; y = 0
∴ From (2) ;
– 1 + log |e| = c
⇒ c = 0
∴ From (2) ;
e– y/x = log |x|
⇒ – $$\frac{y}{x}$$ = log (log |x|)
⇒ y = -x log (log |x|)
which is the required solution.

Question 15.
Solve the differential equation (x2 – y2) dx + 2xy dy = 0, given that y = 1 when x = 1.
Solution:
Given diff. eqn. can be written as

⇒ log (1 + y2) + log x = log C
⇒ log x (1 + y2) = log C
⇒ x $$\left(1+\frac{y^2}{x^2}\right)$$ = C
which is the required general solution.
Given y = 1 when x = 1
∴ from (2) ;
1 + 1 = C
⇒ C = 2
Thus eqn. (2) becomes ;
x2 + y2 = 2x, which is the required solution.

Question 16.
Find particular solutions of the following differential equations:
(i) (x2 + y2) dx = 2xy dy, given that y = 0 when x = 1.
(ii) $$\frac{d y}{d x}=\frac{x y}{x^2+y^2}$$ given that y = 1 when x = 0.
(iii) 2xy + y2 – 2x2 $$\frac{d y}{d x}$$ = 0, given that y (1) = 2.
(iv) (x2 + 3xy + y2) dx – x2 dy = 0 given that y = 0 when x = 1.
(v) $$\frac{d y}{d x}-\frac{y}{x}$$ + cosec ($$\frac{y}{x}$$) = 0, given that y = 0 when x = 1.
(vi) x $$\frac{d y}{d x}$$ = y – x tan $$\left(\frac{y}{x}\right)$$, given that y = $$\frac{\pi}{4}$$ at x = 1.
(vii) x ey/x – y + x $$\frac{d y}{d x}$$ = 0, given that y(e) = 0
Solution:
(i) Given diff. eqn. can be written as ;
$$\frac{d y}{d x}=\frac{x^2+y^2}{2 x y}$$ ………………….(1)
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
Thus eqn. (1) becomes ;

⇒ log |1 – v2| = log |x| – log C
⇒ log |(1 – v2)x| = log C
⇒ (1 – $$\frac{y^2}{x^2}$$) x = A
⇒ x2 – y2 = Ax ……………….(1)
Given y = 0 when x = 1
∴ from (1) ; A = 1
Thus eqn. (1) becomes ;
x2 – y2 = x
which is the required particular solution.

(ii) Given $$\frac{d y}{d x}=\frac{x y}{x^2+y^2}$$ ………………(1)
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
in eqn. (1) ; we have

(iii) Given 2xy + y2 – 2x2 $$\frac{d y}{d x}$$ = 0
⇒ $$\frac{d y}{d x}=\frac{2 x y+y^2}{2 x^2}$$ …………….(1)
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
Using eqn. (2) in eqn. (1) ; we have

(iv) Given diff. eqn. can be written as ;
$$\frac{d y}{d x}=\frac{x^2+3 x y+y^2}{x^2}$$ ……………..(1)
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
in eqn. (1) ; we have

(v) Given $$\frac{d y}{d x}-\frac{y}{x}$$ + cosec ($$\frac{y}{x}$$) = 0
⇒ $$\frac{d y}{d x}$$ = $$\frac{y}{x}$$ – cosec $$\frac{y}{x}$$ ……………..(1)
which is homogeneous diff. eqn.
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
∴ from (1) ; we have
v + x $$\frac{d v}{d x}$$ = v – cosec v
⇒ x $$\frac{d v}{d x}$$ = – cosec v
⇒ $$\frac{d v}{\ {cosec} v}=-\frac{d x}{x}$$ ;
on integrating ; we have
⇒ ∫ sin v dv = – ∫ $$\frac{d x}{x}$$
⇒ – cos v = – log |x| + c
⇒ – cos $$\frac{y}{x}$$ = – log |x| + c ……………….(2)
Since y(1) = 0 i.e. when x= 1 ; y = 0
∴ from (2); we have
– 1 = 0 + c
c = – 1
∴ From (2); we have
1 – cos $$\frac{y}{x}$$ = – log |x|
cos $$\frac{y}{x}$$ – 1= log | x | which is the required solution.

(vi) Given $$\frac{d y}{d x}=\frac{y}{x}-\tan \left(\frac{y}{x}\right)$$ …………………..(1)
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$ in eqn. (1) ;
v + x $$\frac{d v}{d x}$$ = v – tan v
⇒ x $$\frac{d v}{d x}$$ = – tan v
⇒ $$\frac{d v}{\tan v}=-\frac{d x}{x}$$
On integrating both sides
$$\int \frac{d v}{\tan v}=-\int \frac{d x}{x}$$ + log C
⇒ log |sin v| + log |x| = log C
⇒ x sin $$\frac{y}{x}$$ = A …………………..(1)
[∵ A = ± C]
Given y = $$\frac{\pi}{4}$$ at x = 1
∴ from (1) ;
$$\frac{1}{\sqrt{2}}$$ = A
Thus eqn. (1) becomes ;
x tan $$\frac{y}{x}$$ = $$\frac{1}{\sqrt{2}}$$ be the required solution.

(vii) Given x ey/x – y + x $$\frac{d y}{d x}$$ = 0
⇒ $$\frac{d y}{d x}=\frac{y-x e^{y / x}}{x}$$
$$\frac{y}{x}-e^{y / x}$$ …………………(1)
which is homogeneous diff. eqn.
put y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
∴ From (1) ; we have
v + x $$\frac{d v}{d x}$$ = v – ev
⇒ x $$\frac{d v}{d x}$$ = – ev
⇒ e– v dv = – $$\frac{d x}{x}$$ ;
on integrating we have
∫ e– v dv = ∫ – $$\frac{d x}{x}$$
Since y (e) = 0
i.e. when x = e; y = 0
∴ From (2);
– 1 + log |e| = c
⇒ c = 0
∴ From (2) ;
e– y/x = log |x|
⇒ $$-\frac{y}{x}$$ = log (log |x|)
⇒ y = – x log (log |x|)
which is the required solution.

Question 17.
Solve the differential equation x dy – y dx = $$\sqrt{x^2+y^2}$$ dx, given that y = 0 when x = 1.
Solution:
Given diff. eqn. can be written as ;
x dy = (y + $$\sqrt{x^2+y^2}$$) dx
⇒ $$\frac{d y}{d x}=\frac{y+\sqrt{x^2+y^2}}{x}$$ ……………….(1)
putting y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
in eqn. (1) ; we have

Question 18.
Show that the following differential equations are homogeneous and find their particular solutions:
(i) x $$\frac{d y}{d x}$$ sin ($$\frac{y}{x}$$) + x – y sin ($$\frac{y}{x}$$) = 0, given that x = 1 when y = $$\frac{\pi}{2}$$.
(ii) (x ey/x + y) dx = x dy, given that y = 0 when x = 1.
Solution:
(i) Given diff. eqn. be,

(ii) Given diff. eqn. be,
$$\frac{d y}{d x}=\frac{x e^{y / x}+y}{x}$$ ;
putting y = vx
⇒ $$\frac{d y}{d x}$$ = v + x $$\frac{d v}{d x}$$
∴ from (1) ; we get
– e v + x $$\frac{d v}{d x}$$ = $$\frac{x e^v+v x}{x}$$
= ev + v
⇒ x $$\frac{d v}{d x}$$ = $$\frac{x e^v+v x}{x}$$
= ev + v
⇒ x $$\frac{d v}{d x}$$ = ev
⇒ $$\int \frac{d v}{e^v}=\int \frac{d x}{x}$$ + C
⇒ – e-v = log |x| + C
⇒ – e-y/x = log |x| + C ……………..(1)
When x = 1, y = 1
∴ from (1) ; we get
– e-1 = c
c = – $$\frac{1}{e}$$
∴ from (1) ;
– e-y/x = log |x| – $$\frac{1}{e}$$
which is the required solution.