Toronto Math Forum

Question: Find an integrating factor and solve the given equation.
      (3x³y + 2xy +y³) + (x² +y²)y’ = 0

Solution:
let M=3x³y + 2xy +y³, N= x² +y²
M_y=3x³+2x+3y², N_x=2x
since M_y≠N_x —> not exact equation
we need to find the integrationg factor µ(x,y)
multiply µ on both side of the equation
µ(3x³y + 2xy +y³) + µ(x² +y²)y’ = 0
let M¹= µ(3x³y + 2xy +y³) = µM, N¹= µ(x² +y²) = µN
we want M¹_y=N¹_x —>(µM)_y=(µN)_x
µ_yM+µM_y=µ_xN+µN_x
suppose µ is only depend on x
then µ = µ (x), µ_y=0
—>0+µM_y=µ_xN+µN_x
   µ_x/µ= (My-Nx)/N=3
—>µ=e^∫3dx
   =e^3x
we get e^3x(3x³y + 2xy +y³) + e^3x(x² +y²)y’ = 0
M¹= e^3x(3x³y + 2xy +y³), N¹= e^3x(x² +y²)
M¹_y=3x³e^3x+2xe^3x+3y²e^3x,N¹_x=3x³e^3x+2xe^3x+3y²e^3x
M¹_y=N¹_x —> it is exact equation now
By thm, there exist ϕ(x,y) s.t. ϕ_x = M¹, ϕ_y = N¹
ϕ(x,y) = ∫e^3x(3x³y + 2xy +y³)dx
   =x²ye^3x+(1/3)e^3xy³+h(y)
ϕy =x²e^3x+e^3xy²+h’(y)
—>h’(y)=0 —>h(y) is constant
thus, ϕ(x,y)=x²ye^3x+(1/3)e^3xy³=C