Question

xdx+ye^-xdy=0

Answer 1

subtract and seperate ....

Answer 2

If you're expected to treat it as an exact equation:
\[\begin{align*}
x~dx+ye^{-x}~dy&=0\\\\
x+ye^{-x}\frac{dy}{dx}&=0\\\\
M(x,y)+N(x,y)\frac{dy}{dx}&=0
\end{align*}\]
The equation will be exact if \(M_y=N_x\) (partial derivatives with respect to variable in subscript):
\[\frac{\partial M}{\partial y}=0~~\text{and}~~\frac{\partial N}{\partial x}=-ye^{-x}\]
which are not equal.

Integrating factor:
\[\mu(x)=\exp\left(\int\frac{M_y-N_x}{N}~dx\right)=\exp\left(\int dx\right)=e^{x}\]
Distribute the IF:
\[\begin{align*}
xe^{x}+y\frac{dy}{dx}&=0\\\\
M^*(x,y)+N^*(x,y)\frac{dy}{dx}&=0
\end{align*}\]
Now indeed, \(M^*_y=0=N^*_x\), so the equation is exact.

You're looking for a solution of the form \(\Psi(x,y)=C\), and you know that \(\Psi_x=M^*\) and \(\Psi_y=N^*\). You have enough to go on to solve.
\[\begin{align*}\Psi_x&=xe^{-x}\\\\
\int \Psi_x~dx&=\int xe^{-x}~dx\\\\
\Psi&=-xe^{-x}+\int e^{-x}~dx&\text{(integrating by parts)}\\\\
\Psi&=-xe^{-x}-e^{-x}+f(y)&\\\\
\Psi_y&=f'(y)&\\\\
y&=f'(y)\\\\
f(y)&=\frac{1}{2}y^2+C\end{align*}\]