Question

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

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

Answer 1

Answer: (A), (D)

$\sqrt{x^2+y^2}=a\sin \left({\tan }^{-1}y/x\right)+$C
$y=x\tan (const+si{n}^{-1}\frac{1}{a}\sqrt{x^2+y^2})$

Taking $x=r\cos \theta$ and $y=r\sin \theta$
So that $x^2+y^2=r^2$ and $\tan \theta =\frac{y}{x}$
We have $xdx+ydy=rdr$
And $xdy-ydx=x^2{\sec }^2\theta d\theta =r^{2}d\theta$
The given equation can be transformed into
$\frac{rdr}{r^{2}d\theta }=\sqrt{\frac{a^2-r^2}{r^2}}\Rightarrow \frac{dr}{d\theta }=\sqrt{a^2-r^2}$
$\Rightarrow c+{\sin }^{-1}\frac{r}{a}=\theta ={\tan }^{-1}\frac{y}{x}$
$\Rightarrow \sqrt{x^2+y^2}=a\sin (C+{\tan }^{-1}\frac{y}{x})$