Need Solution for R.D. Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 34 Maths Textbook Solution.

Answer: $x\sin \left ( \frac{y}{x} \right )=c\left ( 1+\cos \frac{y}{x} \right )$

Given:$x\frac{dy}{dx}-y+x\sin \left ( \frac{y}{x} \right )=0$

To find: we have to find the solution of given differential equation.

Hint: we will put $y=vx$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}$

Solution: we have,

$x\frac{dy}{dx}-y+x\sin \left ( \frac{y}{x} \right )=0$

$\Rightarrow \frac{dy}{dx}=\frac{y+\sin \frac{y}{x}}{x}$

It is homogeneous equation.

Put $y=vx$ and $\frac{dy}{dx}=v+x\frac{dv}{dx}$

So,

$x \frac{d y}{d x}=\frac{v x-\mathrm{xsin}\left(\frac{y}{x}\right)}{x} \\$

$\Rightarrow x \frac{d v}{d x}= v-\sin v-v \\$

Separating the variables and integrating both side we get

$\int \frac{d v}{\sin v}=-\int \frac{1}{x} d x \\$

$\Rightarrow \int \operatorname{cosec} v d v=-\int \frac{1}{x} d x \\$

$\Rightarrow \log (\operatorname{cosec} v+\cot v)=-\log x+\log c \\$

$\Rightarrow \operatorname{cosec} v+\cot v=\frac{c}{x} \\$

$\Rightarrow \frac{c}{x}=\frac{1}{\operatorname{cosec} v+\cot v} \\$

$\Rightarrow \frac{x}{c}=\frac{1}{\frac{1}{\sin v}+\frac{\cos v}{\sin v}} \\$

$\Rightarrow \frac{x}{c}=\frac{\sin v}{1+\cos v} \\$

$(1+\cos v) x=c \sin v \\$

$\Rightarrow x\left(1+\cos \left(\frac{y}{x}\right)=c \sin \left(\frac{y}{x}\right)\right)$

$\left [ \therefore v=\frac{y}{x} \right ]$
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