Question

$\frac{dy}{dx}=\frac{y}{x}+\sin \left(\frac{y}{x}\right)$

Answer 1

$$\begin{gathered} \frac{dy}{dx}=\frac{y}{x}+\sin \left(\frac{y}{x}\right) \\ \mathrm{This}\mathrm{is}\mathrm{a}\mathrm{homogeneous}\mathrm{differential}\mathrm{equation}. \\ \mathrm{Putting}y=vx\mathrm{and}\frac{dy}{dx}=v+x\frac{dv}{dx},\mathrm{we}\mathrm{get} \\ v+x\frac{dv}{dx}=v+\sin v \\ \Rightarrow x\frac{dv}{dx}=v+\sin v-v \\ \Rightarrow \frac{1}{\sin v}dv=\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{\sin v}dv=\int \frac{1}{x}dx \\ \Rightarrow \int \mathrm{cosec}vdv=\int \frac{1}{x}dx \\ \Rightarrow \log \left|\tan \frac{v}{2}\right|=\log \left|x\right|+\log C \\ \Rightarrow \log \left|\tan \frac{v}{2}\right|=\log \left|x\right|+\log C \\ \Rightarrow \log \left|\tan \frac{v}{2}\right|=\log \left|Cx\right| \\ \Rightarrow \tan \frac{v}{2}=Cx \\ \mathrm{Putting}v=\frac{y}{x},\mathrm{we}\mathrm{get} \\ \Rightarrow \tan \left(\frac{y}{2x}\right)=\mathrm{C}x \\ \mathrm{Hence},\tan \left(\frac{y}{2x}\right)=\mathrm{C}x\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$