Question

The solution of the differential equation $x\frac{dy}{dx}=y+x\tan \frac{y}{x}$, is
(a) $\sin \frac{x}{y}=x+C$

(b) $\sin \frac{y}{x}=Cx$

(c) $\sin \frac{x}{y}=Cy$

(d) $\sin \frac{y}{x}=Cy$

Answer 1

(b) $\sin \frac{y}{x}=Cx$

We have,
$$\begin{gathered} x\frac{dy}{dx}=y+x\tan \frac{y}{x} \\ \Rightarrow \frac{dy}{dx}=\frac{y}{x}+\tan \frac{y}{x}.....\left(1\right) \\ \mathrm{Let}y=vx \\ \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx} \\ \mathrm{Putting}\mathrm{the}\mathrm{above}\mathrm{value}\mathrm{in}\left(1\right),\mathrm{we}\mathrm{get} \\ v+x\frac{dv}{dx}=v+\tan v \\ \Rightarrow x\frac{dv}{dx}=\tan v \\ \Rightarrow \frac{dv}{\tan v}=\frac{dx}{x} \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \log \sin v=\log x+\log C \\ \Rightarrow \log \sin v-\log x=\log C \\ \Rightarrow \log \frac{\sin v}{x}=\log C \\ \Rightarrow \frac{\sin v}{x}=C \\ \Rightarrow \sin v=Cx \\ \Rightarrow \sin \left(\frac{y}{x}\right)=Cx\left[\because y=vx\right] \end{gathered}$$