What is the general solution for y'' + 4y' + 4y = 5xe^(-2x)?

[Squeezebox]

Homework Statement

Find the general solution

y'' + 4y' +4y = 5xe^(-2x)

The Attempt at a Solution

I got (5/2)x^3*e^(-2x) as a particular solution. But I checked online at wolfram alpha and it says the particular solution is (5/6)x^3*e^(-2x). Using method of undetermined coefficients.

 

[rock.freak667]

Solve the homogeneous part of y'' + 4y' +4y = 0. What are the roots of the characteristic equation?

 

[Squeezebox]

Solve the homogeneous part of y'' + 4y' +4y = 0. What are the roots of the characteristic equation?

The homogeneous part gave me

y=c₁e^-2x+c₂xe^-2x

from the root (D+2)².

Multiplying by another (D+2) annihilated the 5xe^-2x, resulting in a new solution

y=c₁e^-2x+c₂xe^-2x+c₃x²e^-2x

is that right?

 

[rock.freak667]

Multiplying by another (D+2) annihilated the 5xe^-2x, resulting in a new solution

y=c₁e^-2x+c₂xe^-2x+c₃x²e^-2x

Your right side is xe^-2x, since -2 is seen twice in the homogeneous part, you must multiply it by x².


Can you show your work on how you got the constant to be 5/2?

 

[Squeezebox]

Your right side is xe^-2x, since -2 is seen twice in the homogeneous part, you must multiply it by x².


Can you show your work on how you got the constant to be 5/2?

(D²+4D+4)(c₃x²e^-2x) = 5xe^-2x

c₃(4x²e^-2x-8xe^-2x+2e^-2x+4(2xe^-2x-2x²e^-2x)+4x²e^-2x) = 5xe^-2x


Every thing but c₃(2e^-2x) reduces to zero

c₃(2e^-2x)=5xe^-2x
c₃=(5/2)x

 

[rock.freak667]

(D²+4D+4)(c₃x²e^-2x) = 5xe^-2x

c₃(4x²e^-2x-8xe^-2x+2e^-2x+4(2xe^-2x-2x²e^-2x)+4x²e^-2x) = 5xe^-2x


Every thing but c₃(2e^-2x) reduces to zero

c₃(2e^-2x)=5xe^-2x
c₃=(5/2)x

I thought you got y_p=c₃x³e^-2x

 

[Squeezebox]

I thought you got y_p=c₃x³e^-2x

y_p=c₃x²e^-2x
y_p=(5/2)x*x²e^-2x
y_p=(5/2)x³e^-2x