Toronto Math Forum
MAT2442019F => MAT244Test & Quizzes => Quiz2 => Topic started by: Juntian Ye on October 04, 2019, 06:17:22 PM

Question: Find an integrating factor and solve the given equation.
(3x^{3}y + 2xy +y^{3}) + (x^{2} +y^{2})y’ = 0
Solution:
let M=3x^{3}y + 2xy +y^{3}, N= x^{2} +y^{2}
M_{y}=3x^{3}+2x+3y^{2}, 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(3x^{3}y + 2xy +y^{3}) + µ(x^{2} +y^{2})y’ = 0
let M^{1}= µ(3x^{3}y + 2xy +y^{3}) = µM, N^{1}= µ(x^{2} +y^{2}) = µN
we want M^{1}_{y}=N^{1}_{x} —>(µM)_{y}=(µN)_{x}
µ_{y}M+µM_{y}=µ_{x}N+µN_{x}
suppose µ is only depend on x
then µ = µ (x), µ_{y}=0
—>0+µM_{y}=µ_{x}N+µN_{x}
µ_{x}/µ= (MyNx)/N=3
—>µ=e^{∫3dx}
=e^{3x}
we get e^{3x}(3x^{3}y + 2xy +y^{3}) + e^{3x}(x^{2} +y^{2})y’ = 0
M^{1}= e^{3x}(3x^{3}y + 2xy +y^{3}), N^{1}= e^{3x}(x^{2} +y^{2})
M^{1}_{y}=3x^{3}e^{3x}+2xe^{3x}+3y^{2}e^{3x},N^{1}_{x}=3x^{3}e^{3x}+2xe^{3x}+3y^{2}e^{3x}
M^{1}_{y}=N^{1}_{x }—> it is exact equation now
By thm, there exist ϕ(x,y) s.t. ϕ_{x }= M^{1}, ϕ_{y} = N^{1}
ϕ(x,y) = ∫e^{3x}(3x^{3}y + 2xy +y^{3})dx
=x^{2}ye^{3x}+(1/3)e^{3x}y^{3}+h(y)
ϕy =x^{2}e^{3x}+e^{3x}y^{2}+h’(y)
—>h’(y)=0 —>h(y) is constant
thus, ϕ(x,y)=x^{2}ye^{3x}+(1/3)e^{3x}y^{3}=C