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Provide Solution For  R.D.Sharma Maths Class 12 Chapter 21 Differential Equations Exercise 21.9 Question 27 Maths Textbook Solution.

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

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

To Find:  We have to find the solution of the given differential equation.

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

Solution: We have,

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

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

It is homogeneous equation.

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


\Rightarrow v+x\frac{dv}{dx}=\frac{vx-x\cos ^{2}\left ( \frac{vx}{x} \right )}{x}

\begin{aligned} &\Rightarrow x \frac{d v}{d x}=v-\cos ^{2} v-v \\ &\Rightarrow x \frac{d v}{d x}=-\cos ^{2} v \\ \end{aligned}

Separating the variables and integrating both sides we get

\Rightarrow \int \frac{d v}{\cos ^{2} v}=-\int \frac{1}{x} d x \\

\Rightarrow \int \sec ^{2} v d v=-\int \frac{1}{x} d x \\

\Rightarrow \tan v= -\log x+\log c \\                                                                        \left [ \therefore \int \sec ^{2}vdv=\tan v \right ]

\Rightarrow \tan \frac{y}{x}=\log \frac{c}{x}                                                                                                                             \left [ \therefore v=\frac{y}{x} \right ]

This is required solution.

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