Have you ever used Variation of Parameters?.
Whew, it's time for beddy-bye
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...
Another method is to use Method of Undetermined Coefficients.
The general solution is
Since the RHS has we look for a solution of the form . But that is among the solutions! So we multiply by to look for a solution of the form . But that is also among the solutions! So we look for a solution of the form .