(x+2)sin(y) + x⋅cos(y)y’= 0. μ =x⋅e^x
Solution:
My =(x+2)cos(y)
Nx = cos(y)
∴My ≠ Nx it is not Exact
Multiply μ
x⋅e^x (x+2)sin(y) + x⋅e^x⋅x⋅cos(y)y’= 0
My = x(x+e)e^x⋅cos(y)
Nx = x(x+e)e^x⋅cos(y)
Now My = Nx and it is Exact
ψy = N ψ = ∫ N ⅾy =∫ x⋅e^x⋅x⋅cos(y) ⅾy
∴ψ = x^2⋅e^x⋅sin(y) + h(x)
∵ψx = M ψx = (2+x)⋅x⋅e^x⋅sin(y) +h’(x) = M
∴h’(x) =0 ∴h(x) = C
ψ(x,y) = x^2⋅e^x⋅sin(y) = C
