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  • Provide Solution For R.D. Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 22 Maths Textbook Solution.

Provide Solution For  R.D. Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 22 Maths Textbook Solution.

Answers (1)

Answer: \tan \left ( \frac{y}{x} \right )=log\left ( \frac{c}{x} \right )

Given:x\frac{dy}{dx}=y-x\cos ^{2}\left ( \frac{x}{y} \right )

To Solve: We have to solve the given differential equation.

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

Solution: Here, given that

\Rightarrow x\frac{dy}{dx}=y-x\cos ^{2}\left ( \frac{x}{y} \right )

\Rightarrow \frac{dy}{dx}=\frac{y-x\cos ^{2}\left ( \frac{y}{x} \right )}{x}\rightarrow \left ( 1 \right )

It is homogeneous equation.

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

So, equation \left ( 1 \right ) becomes

\begin{aligned} &\Rightarrow v+x \frac{d v}{d x}=\frac{v x-x \cos ^{2}\left(\frac{v x}{x}\right)}{x} \\ &\Rightarrow v+x \frac{d v}{d x}=v-\cos ^{2} v \\ &\Rightarrow x \frac{d v}{d x}=-\cos ^{2} v \end{aligned}

Separating the variables we get

\begin{aligned} &\Rightarrow \int \frac{d v}{\cos ^{2} v}=-\int \frac{d x}{x} \\ &\Rightarrow \int \sec ^{2} v d v=-\int \frac{d x}{x} \\ &\Rightarrow \tan v=-\log x+\log c \\ \end{aligned}

                                                                                \qquad \Rightarrow \tan \left(\frac{y}{x}\right)=\log \left(\frac{c}{x}\right)\left[\therefore \log c-\log x=\log \left(\frac{c}{x}\right)\right]

This is required solution

 

Posted by

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