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  • Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 36 Sub Question 9 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 36 Sub Question 9 Maths Textbook Solution.

Answers (1)

Answer: \tan \frac{y}{2x}=\frac{2}{x}

Given:x\frac{dy}{dx}-y+x\sin \left ( \frac{y}{x} \right )=o,y\left ( 2 \right )=x

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 )=o,y\left ( 2 \right )=x

\frac{dy}{dx}=\frac{y}{x}-\sin \left ( \frac{y}{x} \right )            ...(i)

It is homogeneous equation.

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

So,

\begin{aligned} &v+x \frac{d v}{d x}=v-\sin v \\ &\Rightarrow x \frac{d v}{d x}=-\sin v \end{aligned}

Separating the variables and integrating both side we get

\begin{aligned} &\int \frac{d v}{-\sin v}=\int \frac{d x}{x}\\ &\Rightarrow \int \operatorname{cosecv} d v=-\int \frac{d x}{x}\\ &\Rightarrow \log \left|\tan \frac{v}{2}\right|=-\log |x|+\log |c|\\ &\text { Putting } \mathrm{v}=\frac{y}{x}\\ &\Rightarrow\left|\tan \frac{y}{2 x}\right|=\frac{c}{x} \end{aligned}...(ii)

It is given that y\left ( 2 \right )=\pi

Putting y\left ( 2 \right )=\pi ,x=2 in equation (ii) we get

\begin{aligned} &\Rightarrow \tan \left(\frac{\pi}{4}\right)=\frac{c}{2} \\ &\Rightarrow 1=\frac{c}{2} \\ &\Rightarrow c=2 \end{aligned}

Putting value of c in equation (ii) we get

\tan \frac{y}{2x}=\frac{2}{x}

This is required solution.

Posted by

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