Question

Prove that $e^x\sqrt{1-y^2}dx+\frac{y}{x}dy=0$ at $y=1$ and $x=0$

Answer 1

Given,

$e^x\sqrt{1-y^2}dx+\frac{y}{x}dy=0$

$x{e}^x\sqrt{1-y^2}dx=-ydy$

$x{e}^{x}dx=-\frac{y}{\sqrt{1-y^2}}dy$

integrating on both sides, we get,

$\int x{e}^{x}dx=-\int \frac{y}{\sqrt{1-y^2}}dy$

$e^{x}x-e^x=-\sqrt{1-y^2}$

$e^x(x-1)=-\sqrt{1-y^2}$

put $y=1,x=0$

we get,

$e^0(0-e^0)=-\sqrt{1-1}$

$0=0$

Hence proved.