Solving y'' - 4y' + 4y = 2e^(2x+3) by undetermined coefficients

[phantom lancer]

Homework Statement

Using the method of undetermined coefficients, determine the general solution of the following second-order, linear, non-homogenous equations.

y'' - 4y' + 4y = 2e^(2x+3)

Homework Equations

I'm not sure what to do from here...
Also, I'm new here. How do I use the superscript for exponents?

The Attempt at a Solution

r^2 - 4r + 4 = 0
r = 2, so y = C1e^(2x) + C2xe^(2x)

I assume Yp = Ax^(2)e^(2x+3)
so, Yp' = 2Ax^(2)e^(2x+3)
Yp'' = 4Ax^(2)e^(2x+3)

plugging them in the equation: 4Ax^(2)e^(2x+3) - 4(2Ax^(2)e^(2x+3) + 4(Ax^(2)e^(2x+3) = Ax^(2)e^(2x+3)

I get 0 = Ax^(2)e^(2x+3)

From here, I don't know what to do. Please Help.

 

[fresh_42]

The syntax goes $e^{2x+3}$ and $C_1$ for $e^{2x+3}$ and $C_1$.
You might want to have a look at https://www.physicsforums.com/help/latexhelp/

 

[scottdave]

There are the little x² above the typing box. When you have two functions multiplied, you need to use the Product rule to take the derivative.

 

[Orodruin]

In your differentiation of the particular solution you are forgetting the term that comes from differentiating the $x^2$.