Exact Differential Equations and Integrating Factor:
Solve: $ye^{xy}\mathrm dx+(xe^{xy}+2y)\mathrm dy=0$
My Trial:
Seeing the $xy$ term, I think of solving it using the following rule:
If $M\mathrm dx+N\mathrm dy=0$ is of the form $f_1(xy)y\mathrm dx+f_2(xy)x\mathrm dy=0$ and $Mx-Ny \ne 0 $ then integrating factor
$=\dfrac{1}{Mx-Ny}$
However,
When I am trying to take $x$ as common from $N$, I get a term $\dfrac{y}{x}$ and cannot solve the equation.
Which other way can I solve?
Best Answer
Write $$ye^{xy}\mathrm dx+(xe^{xy}+2y)\mathrm dy=0$$ $$e^{xy}(y\mathrm dx+x\mathrm dy)=-2y\mathrm dy$$ $$e^{xy}\mathrm d(xy)=-\mathrm d(y^2)$$ $$\mathrm d\left(e^{xy}\right)=-\mathrm d(y^2)$$ $$e^{xy}=-y^2+C$$