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

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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}


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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