Answer:
Option B
Explanation:
Given,
$\frac{dy}{dx}=\tan\left(\frac{y}{x}\right)+\frac{y}{x}$ ..........................(i)
Clearly , the given different equation is homogeneous
On putting y=Vx
$\frac{dy}{dx}=V+x\frac{dV}{dx}$ in Eq.(i) we get
$V+x\frac{dV}{dx}=\tan V+V$
$\Rightarrow$ $x\frac{dV}{dx}=\tan V$
$\Rightarrow$ $\frac{1}{\tan V}dV=\frac{1}{x}dx$
On integrating both sides, we get
$\int \frac{1}{\tan V}dV=\int\frac{1}{x}dx$
$\Rightarrow$ $\int \cot VdV=\log x+\log c$
$\Rightarrow$ log sin x = log x+ log c
$\Rightarrow$ log sin V= log(xc)
$\Rightarrow$ sin v=xc
$\therefore$ $\sin (\frac{y}{x})=xc$
