Question

$x\frac{dy}{dx}+x{\cos }^2\left(\frac{y}{x}\right)=y$

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ x\frac{dy}{dx}+x{\cos }^2\left(\frac{y}{x}\right)=y \\ \frac{dy}{dx}+{\cos }^2\left(\frac{y}{x}\right)=\frac{y}{x} \\ \Rightarrow \frac{dy}{dx}=\frac{y}{x}-{\cos }^2\left(\frac{y}{x}\right) \\ \mathrm{Putting}y=vx,\mathrm{we}\mathrm{get} \\ \frac{dy}{dx}=v+x\frac{dv}{dx} \\ \therefore v+x\frac{dv}{dx}=v-{\cos }^2\left(v\right) \\ \Rightarrow x\frac{dv}{dx}=-{\cos }^{2}v \\ \Rightarrow {\sec }^{2}vdv=-\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \mathrm{s}\mathrm{e}{\mathrm{c}}^{2}v\mathit{}dv=-\int \frac{1}{x}dx \\ \Rightarrow \tan v\mathit{}=-\log x+C \\ \Rightarrow \tan \frac{y}{x}\mathit{}=-\log x+C \end{gathered}$$