Solution of $\frac{d y}{d x}=\frac{y}{x}+\tan \frac{y}{x}$ is :

$\sin \left(\frac{y}{x}\right)=k x$ $\cos \frac{y}{x}=kx$ $\tan \frac{y}{x}=k x$ none of these

$\sin \left(\frac{y}{x}\right)=k x$

$\frac{d y}{d x}=\frac{y}{x}+\tan \frac{y}{x}$ put $y=vx \Rightarrow v+x \frac{d v}{d x}=v+\tan v$ $\cot v dv=\frac{d x}{x}$ Integrating, we get $\ln \sin v=\ln x+\ln k \Rightarrow \sin \frac{y}{x}=k x$ Hence (1) is the correct answer.