Question

Solution of $\frac{xdx+ydy}{xdy-ydx}=\frac{\sqrt{a^2-x^2-y^2}}{x^2+y^2}$ is

• A. $\sqrt{x^2+y^2}=a\sin \left(C+{\tan }^{-1}\frac{y}{x}\right)$
• B. $\sqrt{x^2+y^2}=a\cos \left(C+{\tan }^{-1}\frac{y}{x}\right)$
• C. $\sqrt{x^2+y^2}=a\tan \left(C+{\tan }^{-1}\frac{y}{x}\right)$
• D. none of these

Answer 1

Answer: (A)

Put $x=rcos\theta ,y=rsin\theta$
$\Rightarrow x^2+y^2=r^2$
$tan\theta =\frac{y}{x}$
$\therefore d\left(x^2+y^2\right)=d\left(r^2\right)$
$\Rightarrow xdx+ydy=rdr$ and $d\left(\frac{y}{x}\right)=d\left(tan\theta \right)$
$\Rightarrow \frac{xdy-ydx}{x^2}=se{c}^2\theta d\theta$
$\therefore$ given diff. eqn. becomes
$\frac{rdr}{r^{2}co{s}^2\theta .se{c}^2\theta d\theta }=\frac{\sqrt{a^2-r^2}}{r^2}$
$\Rightarrow \frac{1}{r}\frac{dr}{d\theta }=\frac{\sqrt{a^2-r^2}}{r}$
$\Rightarrow \frac{dr}{d\theta }=\sqrt{a^2-r^2}$
$\Rightarrow \int \frac{dx}{\sqrt{a^2-r^2}}=\int d\theta +C$
$\Rightarrow si{n}^{-1}\frac{r}{a}=\theta +C$
$\Rightarrow \frac{r}{a}=sin\left(\theta +C\right)$
$\Rightarrow r=asin\left(\theta +C\right)$
$\Rightarrow \sqrt{x^2+y^2}=asin\left[C+ta{n}^{-1}\frac{y}{x}\right]$