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

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

Question

$$\frac{xdx+ydy}{xdy-ydx}=\sqrt{\frac{a^2-x^2-y^2}{x^2+y^2}}$$

solution: $$sin^{-1}(\frac{\sqrt{x^2+y^2}}{a})=tan^{-1}(\frac{y}{x})+c$$

praxer · Answer

@sidsiddhartha please help me with this question.

sidsiddhartha · Answer

hey @praxer how are u? :)

praxer · Answer

I am fine but this question is like eating me up. Please help me out. :)

sidsiddhartha · Answer

os is lagging too much here, wait a bit

sidsiddhartha · Answer

$$x=rcos \theta\y=r \sin \theta$$

sidsiddhartha · Answer

again$$x^2+y^2=r^2\differentiating\x+y.\frac{ dy }{ dx}=r \frac{ dr }{ dx }\x.dx+y.dy=r.dr$$
ok?

praxer · Answer

got it :) than for the denominator substitute y/x = v ?????

sidsiddhartha · Answer

yes again u can write
$$\tan \theta=y/x\so\ \frac{ x.dy-ydx }{ x^2 }=\sec^2 \theta.d \theta\x.dy-y.dx=r^2.d \theta$$

sidsiddhartha · Answer

now just substitute them--
$$\frac{ r.dr }{ r^2.d \theta }=\sqrt{\frac{ 1-r^2 }{ r^2 }}$$

sidsiddhartha · Answer

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

sidsiddhartha · Answer

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

sidsiddhartha · Answer

got it?

praxer · Answer

$$\color{red}{Thank \ You}$$

sidsiddhartha · Answer

yw!

sidsiddhartha · Answer

gotta go now :)
good night !!

praxer · Answer

Good night :) Lord be with you :)