Answer to Question #340399 in Differential Equations for Vee

Question

1/x+ye^xy+2x)dx +(1/y+xe^xy+2y)dy=0

Answer 1

$\left(\frac{1}{x}+ye^{xy}+2x\right)dx+\left(\frac{1}{y}+x{e}^{xy}+2y\right)dy=0$ ,

$P\left(x,y\right)=\frac{1}{x}+y^{xy}+2x, Q\left(x,y\right)=\frac{1}{y}+x{e}^{xy}+2y$ ,

$\frac{\partial P\left(x,y\right)}{\partial y}=e^{xy}+yxe^{xy}$ , $\frac{\partial Q\left(x,y\right)}{\partial x}=e^{xy}+xye^{xy}$ ,

$\frac{\partial P\left(x,y\right)}{\partial y}=\frac{\partial Q\left(x,y\right)}{\partial x}$ - total differential equation

$dU=\frac{\partial U}{\partial x}dx+\frac{\partial U}{\partial y}dy=P\left(x,y\right)dx+Q\left(x,y\right)dy$,

$\int\left(\frac{1}{x}+ye^{xy}+2x\right)dx=\ln{x}+y\frac{1}{y}e^{xy}+x^2+\varphi\left(y\right)=\ln{x}+e^{xy}+x^2+\varphi\left(y\right)$ ,

$\frac{\partial U}{\partial y}=xe^{xy}+\varphi\prime\left(y\right)=\frac{1}{y}+xe^{xy}+2y$ ,

$\varphi\prime\left(y\right)=\frac{1}{y}+2y$ ,

$\varphi\left(y\right)=\int\left(\frac{1}{y}+2y\right)dy=\ln{y}+y^2$ ,

Answer: $U\left(x,y\right)=\ln{x}+e^{xy}+x^2+\ln{y}+y^2=C$