Question

An equation of the curve satisfying $xdy-ydx=\sqrt{x^2-y^2}dx$ and $y\left(1\right)=0$ is

• A. $y=x\sin \left(\log x\right)+C$
• B. $y=x\sin \left(\log |\sin x|\right)$
• C. $y=y^2=x{(x-1)}^2$
• D. $y=2x^2(x-1)$

Answer 1

The correct option is A $y=x\sin \left(\log x\right)+C$

$xdy-ydx=\sqrt{x^2-y^2}dx$

$xdy=ydx+\sqrt{x^2-y^2}dx$

$xdy=(y+\sqrt{x^2-y^2})dx$

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

put $y=vx\Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$

$\therefore v+x\frac{dv}{dx}=\frac{vx+\sqrt{x^2-(vx{)}^2}}{x}$

$v+x\frac{dv}{dx}=\sqrt{1-v^2}+v$

$x\frac{dv}{dx}=\sqrt{1-v^2}$

$\frac{dv}{\sqrt{1-v^2}}=\frac{dx}{x}$

${\sin }^{-1}v=\log x+C$

${\sin }^{-1}\left(\frac{y}{x}\right)=\log x+C$

$\frac{y}{x}=\sin \left(\log x\right)+C$

$y=x\sin \left(\log x\right)+C$