Answer to Question #294528 in Differential Equations for Raima

Question

( x - 2e^y ) dy + y + ( xsin(x) ) dx=0

Answer 1

$$P(x,y)=y+x\sin x, \dfrac{\partial P}{\partial y}=1$$

$$Q(x,y)=x-2e^y, \dfrac{\partial Q}{\partial x}=1$$

$$\dfrac{\partial P}{\partial y}=1=\dfrac{\partial Q}{\partial x}$$

Write the system of two differential equations that define the function $u(x,y)$

$$\dfrac{\partial u}{\partial x}=y+x\sin x$$

$$\dfrac{\partial u}{\partial y}=x-2e^y$$

$$u=\int(x-2e^y)dy+\varphi(x)$$

$$u=xy-2e^y+\varphi(x)$$

$$\dfrac{\partial u}{\partial x}=y+\varphi'(x)=y+x\sin x$$

$$\varphi'(x)=x\sin x$$

$$\varphi(x)=\int x\sin xdx$$

$$\varphi(x)=-x\cos x+\sin x+C$$

$$u=xy-2e^y-x\cos x+\sin x+C$$

$$-xy+2e^y+x\cos x-\sin x=C$$