Question

The solution of the differential equation $x \frac{d y}{d x}=y-x \tan \left(\frac{y}{x}\right)$ is (Here, $k$ is an arbitrary constant )

• A. $x=y \sin ^{-1}\left(\frac{k}{x}\right)$
• B. $y=x \sin ^{-1}\left(\frac{k}{x}\right)$
• C. $x \sin y+k=0$
• D. $y=x \cos (k x)$

Answer 1

Answer: (B)

Given differential equation is
$x \frac{d y}{d x}=y-x \tan \frac{y}{x}$
$\Rightarrow \,\frac{d y}{d x}=\frac{y}{x}-\tan \left(\frac{y}{x}\right)$
Let $ y=v \cdot x$
$\Rightarrow \, \frac{d y}{d x}=v+x \frac{d v}{d x}$
So, $ v+x \frac{d v}{d x}=v-\tan\, v$
$ \Rightarrow \,\frac{d v}{\tan v}=-\frac{d x}{x}$
$\Rightarrow \, \int \cot v d v=\int\left(-\frac{1}{x}\right) d x $
$\Rightarrow \,d \log |\sin v|=-\log |x|+\log k$
$\Rightarrow \,\sin v=\frac{k}{x} $
$\Rightarrow \,v=\sin ^{-1}\left(\frac{k}{x}\right)$
$ \Rightarrow \,y=x \sin ^{-1}\left(\frac{k}{x}\right)$