Question

$\frac{dy}{dx}=\frac{y}{x}-\sqrt{\frac{y^2}{x^2}-1}$

Answer 1

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