Question

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

Answer 1

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

On separating the variables and integrating, we get

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

For RHS, let $I=\int \frac{-y}{\sqrt{1-y^2}}dy$

Let $\sqrt{1-y^2}=z$

So, $\frac{-2y}{\sqrt{1-y^2}}dy=dz$

So, $\frac{-y}{\sqrt{1-y^2}}dy=dz$

So, $z=\int dz=z+c=\sqrt{1-y^2}+c$

For RHS,

Let $J=\int x{e}^{x}dx$

Using product rule, we get

$J=x\int e^{x}dx-\int [\frac{d\left(x\right)}{dx}\int e^{x}dx]dx$

$J+x{e}^x-e^x=e^x(x-1)+c$

So, going to original equation, we get

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