Question

Solve:$x\cos \frac{y}{x}(ydx+xdy)=y\sin \frac{y}{x}(xdy-ydx)$

Answer 1

Given the differential equation,

$x\cos \frac{y}{x}(ydx+xdy)=y\sin \frac{y}{x}(xdy-ydx)\Rightarrow \frac{ydx+xdy}{xdy-ydx}=\frac{y\sin \frac{y}{x}}{x\cos \frac{y}{x}}$

By componendo-dividendo,

$\frac{xdy}{ydx}=\frac{y\sin \frac{y}{x}+x\cos \frac{y}{x}}{y\sin \frac{y}{x}-x\cos \frac{y}{x}}\Rightarrow \frac{x}{y}\frac{dy}{dx}=\frac{x(\frac{y}{x}\sin \frac{y}{x}+\cos \frac{y}{x})}{x(\frac{y}{x}\sin \frac{y}{x}-\cos \frac{y}{x})}\Rightarrow \frac{dy}{dx}=\frac{y}{x}\left\{\frac{(\frac{y}{x}\sin \frac{y}{x}+\cos \frac{y}{x})}{(\frac{y}{x}\sin \frac{y}{x}-\cos \frac{y}{x})}\right\}$

This shows that the given differential equation is homogeneous.

Now, substituting $\frac{y}{x}=v\Rightarrow y=vx\Rightarrow \frac{dy}{dx}=x\frac{dv}{dx}+v$

The differential equation becomes,

$x\frac{dv}{dx}+v=v\left\{\frac{v\sin v+\cos v}{v\sin v-\cos v}\right\}\Rightarrow x\frac{dv}{dx}=v\left\{\frac{v\sin v+\cos v}{v\sin v-\cos v}\right\}-v\Rightarrow x\frac{dv}{dx}=v\{\frac{v\sin v+\cos v}{v\sin v-\cos v}-1\}\Rightarrow x\frac{dv}{dx}=\frac{2v\cos v}{v\sin v-\cos v}\Rightarrow \frac{v\sin v-\cos v}{2v\cos v}dv=\frac{dx}{x}$

Integrating this separable variable type differential equation we get,

$\int \frac{v\sin v-\cos v}{2v\cos v}dv=\int \frac{dx}{x}+\ln c\Rightarrow \frac{1}{2}\int \{\frac{v\sin v}{v\cos v}-\frac{\cos v}{v\cos v}\}dv=\ln x+\ln c\Rightarrow \frac{1}{2}\int (\tan v-\frac{1}{v})dv=\ln \left(cx\right)\Rightarrow \ln \sec v-\ln v=2\ln \left(cx\right)\Rightarrow \ln \left(\frac{\sec v}{v}\right)=\ln {\left(cx\right)}^2\Rightarrow v=\frac{\sec v}{c^2{x}^2}\Rightarrow \frac{y}{x}=\frac{\sec \frac{y}{x}}{c^2{x}^2}\Rightarrow y=\frac{\sec \frac{y}{x}}{c^{2}x}\Rightarrow xy\cos \frac{y}{x}=c^{-2}$

Hence, the required general solution is $xy\cos \frac{y}{x}=p,(p=c^{-2}).$