Question

xdy-ydx=x(x^2-y^2)^1/2

Answer 1

$Given\ differential\ equation \ is \ xdy-ydx=x(\sqrt{x^2-y^2})...(1)\\ Dividing \ the \ equation \ in \ (1) \ with \ \sqrt{x^2-y^2} \, we \ get \\ \frac{x}{\sqrt{x^2-y^2}}\frac{dy}{dx}-\frac{y}{\sqrt{x^2-y^2}}=x\\ Substituting \ y=vx, and \ \frac{dy}{dx}=v+x\frac{dv}{dx}, we \ get \\ \frac{x}{\sqrt{x^2-v^2x^2}}(v+x\frac{dv}{dx})-\frac{vx}{\sqrt{x^2-v^2x^2}}=x\\ \Rightarrow \frac{x}{\sqrt{x^2(1-v^2)}}(v+x\frac{dv}{dx})-\frac{vx}{\sqrt{x^2(1-v^2)}}=x\\ \Rightarrow \frac{x}{x\sqrt{(1-v^2)}}(v+x\frac{dv}{dx})-\frac{vx}{x\sqrt{(1-v^2)}}=x\\ \Rightarrow \frac{1}{\sqrt{(1-v^2)}}(v+x\frac{dv}{dx})-\frac{v}{\sqrt{(1-v^2)}}=x\\ (by \ cancellation)\\ \Rightarrow \frac{v}{\sqrt{(1-v^2)}}+(\frac{dv}{dx})\frac{x}{\sqrt{(1-v^2)}}-\frac{v}{\sqrt{(1-v^2)}}=x\\ \Rightarrow (\frac{dv}{dx})\frac{x}{\sqrt{(1-v^2)}}=x (by \ cancellation \ law)\\ \Rightarrow (\frac{dv}{dx})\frac{1}{\sqrt{(1-v^2)}}=1 (by \ cancellation \ law)\\ Separating \ the \ variables \ v \ and \ x\\ we \ get \ \frac{dv}{\sqrt{(1-v^2)}}=1 dx\\ Integrating \ on \ both \ sides, \ we \ get \\ \int\frac{dv}{\sqrt{(1-v^2)}}=\int 1 dx\\ \Rightarrow sin^{-1}(v)=x+C, where \ C \ is \ a \ constant \ of\ integration\\ Substituting \ back, \ v= \frac{y}{x}, we \ get \\ sin^{-1}(\frac{y}{x})=x+C\\ Taking \ sin \ on \ both \ sides \ we \ get \\ sin(sin^{-1}(\frac{y}{x}))=sin (x+C)\\ \therefore \frac{y}{x}=sin (x+C)$