Toronto Math Forum

Since the equation is not exact, we need to find a function µ(x, y), such that

µ(3x²y + 2xy + y³) + µ(x² + y²)y’ = 0 is exact.

M = µ(3x²y + 2xy + y³)      N = µ(x² + y²)

Then  M_y  = µ_y’(3x²y + 2xy + y³) + µ(3x² + 2x + 3y²)
         N_x = µ_x’(x² + y²) + µ(2x)

Let M_y = N_x , we get µ_y’(3x²y + 2xy + y³) + µ(3x² + 3y²) = µ_x’(x² + y²)

First suppose µ is a function of x only.

Then µ_y’ = 0, we get
µ(3x² + 3y²) = µ_x’(x² + y²)
3µ = ∂µ/∂x

By separable equation,
∫(1/µ) ∂µ = ∫3 ∂x
ln(µ) = 3x
µ = e^3x (the integration factor)

By theorem, there exist ϕ(x,y) , such that , ϕ_x = M, ϕ_y = N

ϕ = ∫e^3x(3x²y + 2xy + y³) = e^3xx²y + (e^3xy³)/3 + h(y)
ϕ_y = e^3xx² + h'(y) = e^3xx² + e^3xy²
Thus, h'(y) = e^3xy², h(y) = (e^3xy³)/3 + C

Therefore, the general solution is ϕ = e^3xx²y + (e^3xy³)/3 = C