Question

Find the particular solution of the differential equation $e^x\sqrt{1-y^2}dx+\frac{y}{x}dy=0$, given that $y=1$ when $x=0$.

Answer 1

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

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

Integrating both sides,

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

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

For $x=0,y=1,$ we get

$\Rightarrow \left(0\right)e^0-e^0=\sqrt{1-(1{)}^2}+C\Rightarrow C=-1$

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