Solve: y'' - 4y' + 4y = 2e^2x .

y'' - 4y' + 4y = 2e^2x

Note that it is a non-homogeneous second order linear equation.

The associated homogeneous equation:
- has constant coefficients
- the non-homogeneous term g(x)=2e^(2x) is in a special form g(x) = P(x)e^(αx)+cos(βx) or g(x) = P(x)e^(αx)+sin(βx). We have P(x)=2, α=2, β=0

Hence we use the method of undetermined coefficient.

The characateristic equation:
r^2 - 4r + 4 = 0
(r - 2)^2 = 0
r = 2

The general solution is given as:
y[g] = Ae^(2x) + Bxe^(2x) where A, B are some constants

The particular solution is given as:
y[p] = C(x^s)e^(αx) where C is a constant
and s counts how many times α+iβ is a root of the character equation.
α+iβ=2, hence s=2.

y[p] = C(x^2)e^(2x)
y[p]' = 2C(x^2+x)e^(2x)
y[p]'' = 2C(2x^2+4x+1)e^(2x)

Put this into our equation:
2C(2x^2+4x+1)e^(2x) - 8Cx(x^2+x)e^(2x) + 4C(x^2)e^(2x) = 2e^2x
C(2x^2+4x+1) - 4C(x^2+x) + 2Cx^2 = 1
C = 1

Hence
y[p] = (x^2)e^(2x)

The general solution is given by:
y = y[g] + y[p]
= Ae^(2x) + Bxe^(2x) + (x^2)e^(2x)

let p(x)=p^2-4p+4
p(x)=0 p=2(repeated)

Yc=Ae^2x+Bxe^2x

Yp=x^2*2e^2x/p''(2)=x^2e^2x

Y=Yc+Yp=Ae^2x+Bxe^2x+x^2e^2x