Question

$\text{The solution of}{\left(\frac{xdy-ydx}{xdx+ydy}\right)}^2=\frac{x^2+y^2}{1-x^2-y^2}\text{is}$

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

Answer 1

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

The correct options are
A $\sqrt{x^2+y^2}=\sin \left[{\tan }^{-1}\right(\frac{y}{x})+c]$
B $\sqrt{x^2+y^2}=\cos \left[{\tan }^{-1}\right(\frac{y}{x})+c]$
D $y=x\tan ({\sin }^{-1}\sqrt{x^2+y^2}+c)$

${\left(\frac{xdy-ydx}{xdx+ydy}\right)}^2=\frac{x^2+y^2}{1-(x^2+y^2)}\Rightarrow \frac{xdx+ydy}{\sqrt{1-(x^2+y^2)}}=\frac{xdy-ydx}{\sqrt{x^2+y^2}}\text{Since,}d\left({\tan }^{-1}\right(\frac{y}{x}\left)\right)=\frac{xdy-ydx}{x^2+y^2}\text{and}d(x^2+y^2)=2(xdx+ydy)\Rightarrow \frac{\frac{1}{2}d(x^2+y^2)}{\left(\sqrt{x^2+y^2}\right)\sqrt{1-(x^2+y^2)}}=\frac{xdy-ydx}{x^2+y^2}=d\left({\tan }^{-1}\right(\frac{y}{x}\left)\right)\text{Now, put}{x}^2+y^2=z^2\text{in the L.H.S., then}\Rightarrow \frac{zdz}{z\sqrt{1-z^2}}=d{\tan }^{-1}\left(\frac{y}{x}\right)\text{Integrating both sides, we get}{\sin }^{-1}z={\tan }^{-1}\left(\frac{y}{x}\right)+c\Rightarrow {\sin }^{-1}\left(\sqrt{x^2+y^2}\right)={\tan }^{-1}\left(\frac{y}{x}\right)+c\Rightarrow \sqrt{x^2+y^2}=\sin \left[{\tan }^{-1}\right(\frac{y}{x})+c]\Rightarrow \sqrt{x^2+y^2}=\cos [\frac{\pi }{2}-{\tan }^{-1}(\frac{y}{x})+c]\Rightarrow \sqrt{x^2+y^2}=\cos \left[{\tan }^{-1}\right(\frac{y}{x})+c]\text{Also}y=x\tan \left[{\sin }^{-1}\right(\sqrt{x^2+y^2})+c]$