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  • Explain Solution R D Sharma Class 12 Chapter 21 Differential Equations Exercise 21 Point 9 Question 36 Sub Question 8 Maths Textbook Solution

Explain Solution R.D.Sharma Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 36 Sub Question 8 Maths Textbook Solution.

Answers (1)

Answer: \cot \left ( \frac{y}{x} \right )=log\mid ex\mid

Given:\left\{x \sin ^{2} \frac{y}{x}-y\right\} d x+x d y=0, y(1)=\frac{\pi}{4}

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,

\left\{x \sin ^{2} \frac{y}{x}-y\right\} d x+x d y=0, y(1)=\frac{\pi}{4}

-\sin ^{2}\frac{y}{x}+\frac{y}{x}=\frac{dy}{dx}                .....(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}=-\sin ^{2} v+v \\ &\Rightarrow x \frac{d v}{d x}=-\sin ^{2} v \end{aligned}

Separating the variables and integrating both side we get

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

It is given that y=\frac{\pi }{4} when x=1

Putting y=\frac{\pi }{4},x=1  in equation (ii) we get

\Rightarrow \cot \left ( \frac{\pi }{4} \right )=log\mid c\mid

\Rightarrow c=e

Putting value of c in equation (ii) we get

\Rightarrow \cot \left ( \frac{\pi }{4} \right )=log\mid ex\mid

This is required solution.

Posted by

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