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Use the method of undetermined coefficients to find the general solution to the de y′′−3y′ 2y=ex e2x e−x

Sagot :

LammettHashidn

I'll assume the ODE is

[tex]y'' - 3y' + 2y = e^x + e^{2x} + e^{-x}[/tex]

Solve the homogeneous ODE,

[tex]y'' - 3y' + 2y = 0[/tex]

The characteristic equation

[tex]r^2 - 3r + 2 = (r - 1) (r - 2) = 0[/tex]

has roots at [tex]r=1[/tex] and [tex]r=2[/tex]. Then the characteristic solution is

[tex]y = C_1 e^x + C_2 e^{2x}[/tex]

For nonhomogeneous ODE (1),

[tex]y'' - 3y' + 2y = e^x[/tex]

consider the ansatz particular solution

[tex]y = axe^x \implies y' = a(x+1) e^x \implies y'' = a(x+2) e^x[/tex]

Substituting this into (1) gives

[tex]a(x+2) e^x - 3 a (x+1) e^x + 2ax e^x = e^x \implies a = -1[/tex]

For the nonhomogeneous ODE (2),

[tex]y'' - 3y' + 2y = e^{2x}[/tex]

take the ansatz

[tex]y = bxe^{2x} \implies y' = b(2x+1) e^{2x} \implies y'' = b(4x+4) e^{2x}[/tex]

Substitute (2) into the ODE to get

[tex]b(4x+4) e^{2x} - 3b(2x+1)e^{2x} + 2bxe^{2x} = e^{2x} \implies b=1[/tex]

Lastly, for the nonhomogeneous ODE (3)

[tex]y'' - 3y' + 2y = e^{-x}[/tex]

take the ansatz

[tex]y = ce^{-x} \implies y' = -ce^{-x} \implies y'' = ce^{-x}[/tex]

and solve for [tex]c[/tex].

[tex]ce^{-x} + 3ce^{-x} + 2ce^{-x} = e^{-x} \implies c = \dfrac16[/tex]

Then the general solution to the ODE is

[tex]\boxed{y = C_1 e^x + C_2 e^{2x} - xe^x + xe^{2x} + \dfrac16 e^{-x}}[/tex]

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