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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.

Answers (1)

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 $\frac{dy}{dx}=v+x\frac{dv}{dx}$

So,

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

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

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.

Posted by

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