Question

The solution of $\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{1-x^2-y^2}{x^2+y^2}}$ is

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

Answer 1

Answer: (A), (B), (D)

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

$\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{1-x^2-y^2}{x^2+y^2}}$

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

Since, $d\left({\tan }^{-1}\left(\frac{y}{x}\right)\right)=\frac{xdy-ydx}{x^2+y^2}$ and $d(x^2+y^2)=2(xdx+ydy)$

$\Rightarrow \frac{\frac{1}{2}d(x^2+y^2)}{\left(\sqrt{x^2+y^2}\right)\sqrt{1-(x^2+y^2)}}=\frac{xdy-ydx}{x^2+y^2}=d\left({\tan }^{-1}\right(\frac{y}{x}\left)\right)$

Now, put $x^2+y^2=z^2$ in the L.H.S., then,
$\Rightarrow \frac{zdz}{z\sqrt{1-z^2}}=d{\tan }^{-1}\left(\frac{y}{x}\right)$

Integrating both sides, we get
${\sin }^{-1}z={\tan }^{-1}\left(\frac{y}{x}\right)+c\Rightarrow {\sin }^{-1}\left(\sqrt{x^2+y^2}\right)={\tan }^{-1}\left(\frac{y}{x}\right)+c$
$\Rightarrow \sqrt{x^2+y^2}=\sin [{\tan }^{-1}\left(\frac{y}{x}\right)+c]$
$\Rightarrow \sqrt{x^2+y^2}=\cos [\frac{\pi }{2}-{\tan }^{-1}\left(\frac{y}{x}\right)+c]$
$\Rightarrow \sqrt{x^2+y^2}=\cos [{\tan }^{-1}\left(\frac{y}{x}\right)+C]$

Also, $y=x\tan \left[{\sin }^{-1}\right(\sqrt{x^2+y^2})+c]$