Question

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

Answer 1

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