Answer to Question #294530 in Differential Equations for Raima

Question

Solve the exact differential equation = ( 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 $u(x,y)$

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

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

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

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

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

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

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

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

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

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