Question

The solution of the differential equation $x\frac{dy}{dx}=y-x\tan \left(\frac{y}{x}\right)$ 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. $y=x\cos (kx)$

Answer 1

Answer: (B)

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