Question

Determine the general solution to the exact differential equation

(e^2y-ycosxy)dx+(2xe^2y-xcosxy+2y)dy=0

Answer 1

$$\dfrac{\partial M}{\partial y}=2e^{2y}-\cos(xy)+xy\sin(xy)$$

$$\dfrac{\partial N}{\partial x}=2e^{2y}-\cos(xy)+xy\sin(xy)$$

$$\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}$$

$$\dfrac{\partial u}{\partial x}=e^{2y}-y\cos(xy)$$

$$\dfrac{\partial u}{\partial y}=2xe^{2y}-x\cos(xy)+2y$$

$$u(x,y)=\int(e^{2y}-y\cos(xy))dx+\varphi(y)$$

$$=xe^{2y}-\sin(xy)+\varphi(y)$$

$$\dfrac{\partial u}{\partial y}=2xe^{2y}-x\cos(xy)+\varphi'(y)$$

$$=2xe^{2y}-x\cos(xy)+2y$$

$$\varphi'(y)=2y$$

$$\varphi(y)=y^2+C_1$$

$$u=xe^{2y}-\sin(xy)+y^2+C_1$$

$$xe^{2y}-\sin(xy)+y^2=C$$