Question

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

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ x\frac{dy}{dx}-y=2\sqrt{y^2-x^2} \\ \Rightarrow \frac{dy}{dx}=\frac{2\sqrt{y^2-x^2}+y}{x} \\ \mathrm{This}\mathrm{is}\mathrm{a}\mathrm{homogeneous}\mathrm{differential}\mathrm{equati}\mathrm{on}. \\ \mathrm{Putting}y=vx\mathrm{and}\frac{dy}{dx}=v+x\frac{dv}{dx},\mathrm{we}\mathrm{get} \\ v+x\frac{dv}{dx}=\frac{2\sqrt{v^2{x}^2-x^2}+vx}{x} \\ \Rightarrow v+x\frac{dv}{dx}=2\sqrt{v^2-1}+v \\ \Rightarrow x\frac{dv}{dx}=2\sqrt{v^2-1}+v-v \\ \Rightarrow x\frac{dv}{dx}=2\sqrt{v^2-1} \\ \Rightarrow \frac{1}{2\sqrt{v^2-1}}dv=\frac{1}{x}dx \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \int \frac{1}{2\sqrt{v^2-1}}dv=\int \frac{1}{x}dx \\ \Rightarrow \int \frac{1}{\sqrt{v^2-1}}dv=2\int \frac{1}{x}dx \\ \Rightarrow \log \left|v+\sqrt{v^2-1}\right|=2\log \left|x\right|+\log C \\ \Rightarrow v+\sqrt{v^2-1}=\mathrm{C}{x}^2 \\ \mathrm{Putting}v=\frac{y}{x},\mathrm{we}\mathrm{get} \\ \therefore \frac{y}{x}+\sqrt{\frac{y^2}{x^2}-1}=\mathrm{C}{x}^2 \\ \Rightarrow y+\sqrt{y^2-x^2}=\mathrm{C}{x}^3 \\ \mathrm{Hence},y+\sqrt{y^2-x^2}=\mathrm{C}{x}^3\mathrm{is}\mathrm{the}\mathrm{required}\mathrm{solution}. \end{gathered}$$