If $y=f(x)$ is solution of differential equation $ydx+dy= {-e^{x}}{y^2}dy$ with $y(0)=1$
find the area bounded by curves: $y=e^{x}$,$y=f(x)$&$x=1$
My attempt : since given differential equation is not exact, hence I find it, in fact this is $$u(x,y)=\dfrac{e^{-x}}{y^2}$$ And its solution is : $$y^2=e^{-x}$$ Without seeing their graphs, I have no idea, how to find it. Please give your valuable hints. Also easy way to find solution of differential equation. Thanks in advance.
Since $e^x$ is positive for all real $x$, you can discard the negative solution of $y^2=e^{-x}$ which leaves you with $y = \sqrt{e^{-x}}$. Furthermore, $e^x$ and your differential equation meet at $x=0$ since $y(0)= 1$ and $e^0 = 1$.
So you have to find:
$$\int_0^1 e^x - \sqrt{e^{-x}} \ \mathrm{d} x$$