Question

Solution of $\frac{xdx+ydy}{xdy-ydx}=\frac{\sqrt{1-(x^2+y^2)}}{\sqrt{(x^2+y^2)}}$ is:

• A. ${\sin }^{-1}\sqrt{(x^2+y^2)}=c$
• B. ${\tan }^{-1}\frac{y}{x}=c$
• C. ${\sin }^{-1}\sqrt{(x^2+y^2)}={\tan }^{-1}\frac{y}{x}+c$
• D. None of these

Answer 1

Answer: (C)

${\sin }^{-1}\sqrt{(x^2+y^2)}={\tan }^{-1}\frac{y}{x}+c$

Substitute $x=r\cos \theta ,y=r\sin \theta$
$\therefore dx=-r\sin \theta d\theta +dr\cos \theta dy=r\cos \theta d\theta +dr\sin \theta$

$\therefore$ We have $\frac{r\cos \theta (-r\sin \theta d\theta +dr\cos \theta )+r\sin \theta (r\cos \theta d\theta +dr\sin \theta )}{\left(r\cos \theta \right)(r\cos \theta d\theta +dr\sin \theta )-\left(r\sin \theta \right)(-r\sin \theta d\theta +dr\cos \theta )}=\frac{\sqrt{1-r^2}}{\sqrt{r^2}}$

$\Rightarrow \frac{-r^2\sin \theta \cos \theta d\theta +r{\cos }^2\theta dr+r^2\sin \theta \cos \theta d\theta +r{\sin }^2\theta dr}{r^2{\cos }^2\theta d\theta +r\sin \theta \cos \theta dr+r^2{\sin }^2\theta d\theta -r\sin \theta \cos \theta dr}=\frac{\sqrt{1-r^2}}{\sqrt{r^2}}$

$\Rightarrow \frac{rdr}{r^{2}d\theta }=\frac{\sqrt{1-r^2}}{\sqrt{r^2}}\Rightarrow \frac{dr}{\sqrt{1-r^2}}=d\theta$

$\Rightarrow {\sin }^{-1}r=\theta +c\Rightarrow {\sin }^{-1}\sqrt{x^2+y^2}={\tan }^{-1}\frac{y}{x}+c$