Question: (x+2)* sin(y) + x*cox(y)*(dy/dx) = 0, u(x,y) = x*e^(x)
Solution:M(x,y) = (x+2)*sin(y),N(x,y) = cos(y)
My = (x+2)*cos(y), Nx= cos(y)
Because My is not equal to Nx, the equation is not exact.
In this way, (x+2)*(x*e^(x))*sin(y)+ x^2 *(e^x)*cos(y)*(dy/dx)=0
so, Lx(x,y) = (x+2)x(e^x)sin(y)
Ly(x,y) = (x^2)(e^x)cos(y)
L(x,y) = int((x+2)x*(e^x)sin(y)dx = (x^2)^(e^x)sin(y)+h(y)
Ly(x,y)= (x^2)(e^x)cos(y)+h'(y)= (x^2)(e^x)cos(y)
h'(y) = 0
As a result, (x^2)*(e^x)sin(y)=C
