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

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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.

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