Question

$x\frac{dy}{dx}-y+x\sin \left(\frac{y}{x}\right)=0$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ x\frac{dy}{dx}-y+x\sin \left(\frac{y}{x}\right)=0 \\ \Rightarrow \frac{dy}{dx}=\frac{y-x\sin \left({\displaystyle \frac{y}{x}}\right)}{x} \\ \mathrm{This}\mathrm{is}\mathrm{a}\mathrm{homogenoeus}\mathrm{differential}\mathrm{equation}. \\ \mathrm{Putting}y=vx\mathrm{and}\frac{dy}{dx}=v+x\frac{dv}{dx},\mathrm{we}\mathrm{get} \\ v+x\frac{dv}{dx}=\frac{vx-x\sin v}{x} \\ \Rightarrow x\frac{dv}{dx}=v-\sin v-v \\ \Rightarrow x\frac{dv}{dx}=-\sin v \\ \Rightarrow \mathrm{cosec}vdv=-\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \mathrm{cosec}vdv=-\int \frac{1}{x}dx \\ \Rightarrow -\int \mathrm{cosec}vdv=\int \frac{1}{x}dx \\ \Rightarrow -\log \left|\mathrm{cosec}v-\cot v\right|=\log \left|x\right|+\log C \\ \Rightarrow \log \left|\frac{1}{\mathrm{cosec}v-\cot v}\right|=\log \left|Cx\right| \\ \Rightarrow \log \left|\mathrm{cosec}v+\cot v\right|=\log \left|Cx\right| \\ \Rightarrow \log \left|\frac{1+\cos v}{\sin v}\right|=\log \left|Cx\right| \\ \Rightarrow \frac{1+\cos v}{\sin v}=Cx \\ \Rightarrow x\sin v=\frac{1}{C}\left(1+\cos v\right) \\ \Rightarrow x\sin v=K\left(1+\cos v\right)\left(\mathrm{where},K=\frac{1}{C}\right) \\ \mathrm{Putting}v=\frac{y}{x},\mathrm{we}\mathrm{get} \\ \Rightarrow x\sin \left(\frac{y}{x}\right)=K\left[1+\cos \left(\frac{y}{x}\right)\right] \\ \mathrm{Hence},x\sin \left(\frac{y}{x}\right)=K\left[1+\cos \left(\frac{y}{x}\right)\right]\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$