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

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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 e^x/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| = e^x/y e^c
⇒ x = ± e^c e^x/y
= A e^x/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 + e^y/x) dy + e^y/x (1 – \(\frac{y}{x}\)) dx = 0
(ii) 2ye^x/y dx – (y – 2xe^x/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 + e^v| = – log |x| + log C
[∵ ∫ \(\frac{f^{\prime}(x) d x}{f(d)}\) = log |f(x)| + C]
⇒ log |(v + e^v)x| = log C
⇒ log |y + x e^y/x| = log C
⇒ y + x e^y/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) xe^y/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 xe^y/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 – e^v
⇒ x \(\frac{d v}{d x}\) = – e^v
⇒ 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 e^y/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 e^y/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 – e^v
⇒ x \(\frac{d v}{d x}\) = – e^v
⇒ 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 e^y/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 \(\frac{d v}{d x}\) = \(\frac{x e^v+v x}{x}\)
= e^v + v
⇒ x \(\frac{d v}{d x}\) = \(\frac{x e^v+v x}{x}\)
= e^v + v
⇒ x \(\frac{d v}{d x}\) = e^v
⇒ \(\int \frac{d v}{e^v}=\int \frac{d x}{x}\) + 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.