Differential equations, please check my answers

Ok, just wanted to quickly see if somebody could give a couple of my answers a quick check and point me in the right direction for one question.

The questions are:

Find the general solution of each of the following differential equations.

a) y'' - y = exp(2x) (incase it isn't showing up properly the first term, y'' is d^2y/dx^2)

b) y'' - y = x^2

c) y'' - y = 2exp(x)

And here is what I've done.

I know that for all three to get the general solution you need to find the complentary function and a particular integral, and then the general solution is simply the sum of the two.

a) For the complementary function we let y'' - y = 0, and then we need to figure out the roots of the equation so we take m^2 - 1 = 0 and this gives us m = 1 or m = -1
Then from the roots we know that the complementary function is Aexp(x) + Bexp(-x)

Then we move onto the particular integral. So we need to take a trial function which we define as aexp(2x) and so we now need to find A, so we differentiate it to get y'' = 4aexp(2x) and then substitue this into the first equation to give 4aexp(2x) - aexp(2x) = exp(2x) and from that we find that a = 1/3 and so the particular integral is 1/3exp(2x)

As a result of this the general solution is Aexp(x) + Bexp(-x) + 1/3exp(2x)

b) As the left hand side is the same as in part a we already know that the complementary function is Aexp(x) + Bexp(-x)

So, all we need to find is the particular integral. We take the trial function to be ax^2 + b and differentiating it gives us y'' = 2a, which we then substitue back in to the first equation to give us 2a - (ax^2 + b) = x^2 which we can work through to give a = -1 and b = -2 so when we put it all together we get:

Aexp(x) + Bexp(-x) - x^2 - 2

c) This is the one that is completely stumping me. We have the same complementary function as before. Then the trial function is 2aexp(x) which we differentiate to get y'' = 2aexp(x) but this then gives me the problem that when we substitute it back in we get 2aexp(x) - 2aexp(x) = 2exp(x) and so the left side becomes 0 and it all goes wrong.


Can somebody please give part a & b a quick check for me please, and let me know where I'm going wrong with the third part.

Thanks very much.