Question

Find the solution of $\frac{xdx+ydy}{xdy-dx}=\sqrt{\left(\frac{a^2-x^2-y^2}{x^2+y^2}\right)}.$

• A. ${\sin }^{-1}\frac{\sqrt{x^2+y^2}}{a}={\tan }^{-1}\frac{x}{y}+c$
• B. ${\sin }^{-1}\frac{\sqrt{x^2+y^2}}{a}={\tan }^{-1}\frac{y}{x}+c$
• C. ${\sin }^{-1}\frac{\sqrt{x^3+y^2}}{a}={\tan }^{-1}\frac{x}{y}+c$
• D. ${\sin }^{-1}\frac{\sqrt{x^3+y^2}}{a}={\tan }^{-1}\frac{y}{x}+c$

Answer 1

The correct option is B ${\sin }^{-1}\frac{\sqrt{x^2+y^2}}{a}={\tan }^{-1}\frac{y}{x}+c$

We know that $xdx+ydy=\frac{1}{2}d(x^2+y^2)$ and $\frac{xdy-ydx}{x^2}=d\left(\frac{y}{x}\right)$
The given equation can be written as , $\frac{1}{2}\frac{d(x^2+y^2)}{\sqrt{a^2-(x^2+y^2)}}=(xdy-ydx)\frac{1}{\sqrt{x^2+y^2}}$
or $\frac{1}{2}\frac{d(x^2+y^2)}{\sqrt{x^2+y^2}\sqrt{a^2-(x^2+y^2)}}=\frac{xdy-ydx}{x^2(1+y^2/x^2)}=\frac{d\left(y/x\right)}{(1+y^2/x^2)}$
Put $x^2+y^2=t^2$ for L.H.S.and $y/x=z$
for R.H.S
$\Rightarrow \frac{1}{2}.\frac{2tdt}{t\sqrt{a^2-t^2}}=\frac{1}{1+z^2}\left(dz\right)$
Integrating we get,
${\sin }^{-1}\left(\frac{t}{a}\right)={\tan }^{-1}z+c$
${\sin }^{-1}\frac{\sqrt{x^2+y^2}}{a}={\tan }^{-1}\frac{y}{x}+c$