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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.

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

Answers (1)

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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