Question

$\frac{dy}{dx}=\frac{y^2-x^2}{2xy}$

Answer 1

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