Find the particular solution of the differential equation: y" - 4y' + 4y = e^(2x)?

I'm not sure what to do b/c I get 0 for A?? Can someone help, this is what I had so far:

Y(x) = Ae^2x

Y'(x) = 2Ae^2x

Y"(x) = 4Ae^2x

So... 4Ae^2x - 8Ae^2x + 4Ae^2x = e^2x

Which becomes 0 = e^2x

and then I'm stumped...