Toronto Math Forum
MAT2442018F => MAT244Tests => Quiz2 => Topic started by: Victor Ivrii on October 05, 2018, 05:11:58 PM

Find an integrating factor and solve the given equation.
$$
(3x^2y + 2xy + y^3) + (x^2 + y^2)y' = 0.
$$

Since the equation is not exact, we need to find a function µ(x, y), such that
µ(3x^{2}y + 2xy + y^{3}) + µ(x^{2} + y^{2})y’ = 0 is exact.
M = µ(3x^{2}y + 2xy + y^{3}) N = µ(x^{2} + y^{2})
Then M_{y} = µ_{y}’(3x^{2}y + 2xy + y^{3}) + µ(3x^{2} + 2x + 3y^{2})
N_{x} = µ_{x}’(x^{2} + y^{2}) + µ(2x)
Let M_{y} = N_{x} , we get µ_{y}’(3x^{2}y + 2xy + y^{3}) + µ(3x^{2} + 3y^{2}) = µ_{x}’(x^{2} + y^{2})
First suppose µ is a function of x only.
Then µ_{y}’ = 0, we get
µ(3x^{2} + 3y^{2}) = µ_{x}’(x^{2} + y^{2})
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}(3x^{2}y + 2xy + y^{3}) = e^{3x}x^{2}y + (e^{3x}y^{3})/3 + h(y)
ϕ_{y} = e^{3x}x^{2} + h'(y) = e^{3x}x^{2} + e^{3x}y^{2}
Thus, h'(y) = e^{3x}y^{2}, h(y) = (e^{3x}y^{3})/3 + C
Therefore, the general solution is ϕ = e^{3x}x^{2}y + (e^{3x}y^{3})/3 = C

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