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 $x = r\,cos\,\theta, y = r\,sin\,\theta$
$\Rightarrow x^{2} + y^{2} = r^{2}$
$tan\,\theta=\frac{y}{x}$
$\therefore d\left(x^{2}+y^{2}\right)=d\left(r^{2}\right)$
$\Rightarrow x\,dx + y\,dy = rdr$ and $d\,\left(\frac{y}{x}\right) = d\,\left(tan\,\theta\right)$
$\Rightarrow \frac{xdy-ydx}{x^{2}} = sec^{2}\,\theta\,d\,\theta$
$\therefore $ given diff. eqn. becomes
$\frac{rdr}{r^{2}\,cos^{2}\,\theta. sec^{2}\,\theta\,d\theta} = \frac{\sqrt{a^{2}-r^{2}}}{r^{2}}$
$\Rightarrow \frac{1}{r} \frac{dr}{d\theta} = \frac{\sqrt{a^{2}-r^{2}}}{r}$
$\Rightarrow \frac{dr}{d\theta } = \sqrt{a^{2}-r^{2}}$
$\Rightarrow \int \frac{dx}{ \sqrt{a^{2}-r^{2}}} = \int d\theta + C$
$\Rightarrow sin^{-1} \frac{r}{a} = \theta+C$
$\Rightarrow \frac{r}{a} = sin \left(\theta +C\right)$
$\Rightarrow r = a\,sin \left(\theta +C\right)$
$\Rightarrow \sqrt{x^{2}+y^{2}} = a\,sin\left[C+tan^{-1} \frac{y}{x}\right]$