Same as y"-6y'+9y=7e-2x.
First solve y"-6y'+9y=0 to find the characteristic solution yc.
Use r2-6r+9=0=(r-3)2⇒yc=Ae3x+Bxe3x.
CHECK: yc'=3Ae3x+3Bxe3x+Be3x; yc"=9Ae3x+9Bxe3x+3Be3x+3Be3x=9Ae3x+9Bxe3x+6Be3x.
yc"-6yc'+9yc=9Ae3x+9Bxe3x+6Be3x-6(3Ae3x+3Bxe3x+Be3x)+9(Ae3x+Bxe3x)=0, because all terms cancel.
The particular solution has to involve e-2x so let yp=ae-2x, then yp'=-2ae-2x and yp"=4ae-2x.
yp"-6yp'+9yp=4ae-2x+12ae-2x+9ae-2x=7e-2x, so 25a=7, a=7/25, yp=7e-2x/25.
The solution is y=yc+yp=Ae3x+Bxe3x+7e-2x/25. (A and B are arbitrary constants.)