The solution of $x^2 \frac{d y}{d x}-x y=1+\cos \frac{y}{x}$ is :

$\tan \frac{y}{2 x}=c-\frac{1}{2 x^2}$

$\frac{d y}{d x}-\frac{1}{x}y=\frac{1}{x^2}+\frac{1}{x^2} \cos \frac{y}{x}$             .......(1)

Put y = vx ⇒ $\frac{d v}{d x}-v+x \frac{d v}{d x}$

∴ (1) becomes $v+x \frac{d v}{d x}-v=\frac{1}{x^2}+\frac{1}{x^2} \cos v$

$\Rightarrow x^3 \frac{d v}{d x}=1+\cos v$

$\Rightarrow \frac{d v}{1+\cos v}=\frac{d x}{x^3} \Rightarrow \int \frac{1}{2} \sec ^2 \frac{v}{2} d v=\frac{x^{-2}}{-2}+c$

$\Rightarrow \tan \frac{v}{2}=-\frac{1}{2 x^2}+c$

Hence (1) is the correct answer.