Question

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

Answer 1

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