#### Provide Solution For  R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 19 Maths Textbook Solution.

Answer: $\tan \left ( \frac{y}{2x} \right )=cx$

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

To find: We have to solve the given differential equation.

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

Solution: We have,

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

It is a homogeneous equation

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

So,$v+x\frac{dv}{dx}=v+\sin v$

\begin{aligned} &\Rightarrow x \frac{d v}{d x}=\sin v \\ &\Rightarrow \int \operatorname{cosecv} d v=\int \frac{d x}{x} \\ &\Rightarrow \log \left|\tan \left(\frac{v}{2}\right)\right|=\log x+\log c \\ &\Rightarrow \tan \frac{v}{2}=c x \\ &\Rightarrow \tan \left(\frac{y}{2 x}\right)=c x \end{aligned}                                                                                    $\left [ \therefore v=\frac{y}{x} \right ]$

This is required solution.