Question

2y^2+4x y dy/dx + xe^xy dy/dx + 2y dy/dx=-ye^xy

Answer 1

$2y^2+4x y\frac{ dy}{dx} + xe^{xy} \frac{dy}{dx} + 2y\frac{ dy}{dx}=-ye^{xy}$

$(4x y + xe^{xy} + 2y)\frac{ dy}{dx}+2y^2+ye^{xy}=0$

$(2y^2+ye^{xy})dx+(4x y + xe^{xy} + 2y) dy=0$

$\frac{\partial}{\partial x}(4x y + xe^{xy} + 2y)=4y+e^{xy}+xye^{xy}$

$\frac{\partial}{\partial y}( 2y^2 + ye^{xy})=4y+e^{xy}+xye^{xy}$

Since $\frac{\partial}{\partial x}(4x y + xe^{xy} + 2y)=\frac{\partial}{\partial y}( 2y^2 + ye^{xy})$, there exists a function $U=U(x,y)$ such that $dU=(4x y + xe^{xy} + 2y) dy+(2y^2+ye^{xy})dx.$ Therefore, $\frac{\partial U}{\partial x}=2y^2 + ye^{xy}$ and $\frac{\partial U}{\partial y}=4x y + xe^{xy} + 2y$.

$U(x,y)=\int(2y^2 + ye^{xy})dx=2y^2x+e^{xy}+C(y)$.

$4xy + xe^{xy}+2y=\frac{\partial U}{\partial y}=\frac{\partial }{\partial y}(2y^2x+e^{xy}+C(y))=4xy+xe^{xy}+C'(y)$

Therefore, $C'(y)=2y$. So $C(y)=y^2+C$.

We conclude that the general solution of the differential equation is the following:

$2y^2x+e^{xy}+y^2=C$