This can be written: y"-4y'+4y=3e2x.
First find the characteristic equation by solving r2-4r+4=0=(r-2)2.
Since there's only one root r=2 then the solution is yc=Ae2x+xBe2x, where yc denotes the characteristic solution. Let's check this out:
yc'=2Ae2x+Be2x+2xBe2x; yc"=4Ae2x+4Be2x+4xBe2x;
yc"-4yc'+4yc=
[4Ae2x+4Be2x+4xBe2x]-4[2Ae2x+Be2x+2xBe2x]+4[Ae2x+xBe2x]=
4Ae2x+4Be2x+4xBe2x-8Ae2x-4Be2x-8xBe2x+4Ae2x+4xBe2x=
4Ae2x+4Ae2x-8Ae2x+4Be2x-4Be2x+4xBe2x+4xBe2x-8xBe2x=0✔️
Now we need the particular solution yp. Note that 3e2x is another e2x term. We've already used e2x and xe2x, so it would futile to use either of these.
However, ax2e2x (a is another constant which can be calculated) hasn't been used and, furthermore, the second derivative of ax2e2x would add the extraneous term 2ae2x and therefore would not cancel out the way that e2x and xe2x terms did when deriving yc.
Therefore 2ae2x=3e2x, making a=3/2.
y=yc+yp=Ae2x+xBe2x+(3/2)x2e2x. This can be written y=(A+Bx+(3/2)x2)e2x.