Question

The general solution of the differential equation $x(xdx-ydy)=4\sqrt{x^2-y^2}(xdy-ydx)$ is
(where $c$ is constant of integration)

• A. $\sqrt{x^2-y^2}=\left(\frac{y}{x}\right)+c$
• B. $\ln |\sqrt{x^2-y^2}|=4{\sin }^{-1}\left(\frac{y}{x}\right)+c$
• C. $\sqrt{x^2-y^2}=4{\sin }^{-1}\left(\frac{y}{x}\right)+c$
• D. $\ln |\sqrt{x^2-y^2}|=\left(\frac{y}{x}\right)+c$

Answer 1

The correct option is B $\ln |\sqrt{x^2-y^2}|=4{\sin }^{-1}\left(\frac{y}{x}\right)+c$

$x(xdx-ydy)=4\sqrt{x^2-y^2}(xdy-ydx)\cdots \left(i\right)$
Put $x=r\sec \theta ,y=r\tan \theta$
$\Rightarrow x^2-y^2=r^2$
$\Rightarrow xdx-ydy=rdr$ and
$xdy-ydx=\left(r^2\sec \theta \right)d\theta$
$\therefore$ From $\left(i\right)$
$\Rightarrow \left(r\sec \theta \right)\left(rdr\right)=4r(r^2\cdot \sec \theta d\theta )$
$\Rightarrow \frac{dr}{r}=4d\theta$
Integrating both sides, we get
$\int \frac{dr}{r}=\int 4d\theta$
$\Rightarrow \ln |r|=4\theta +c$
$\therefore \ln |\sqrt{x^2-y^2}|=4{\sin }^{-1}\left(\frac{y}{x}\right)+c$