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 Quote by mathwonk outstanding! and do you know the solution if a = b?
Hey thanks for the help! Yea I just tried it now, and it didn't take me as long as that first one because it's almost the same. There are two differences in the proof. Originally I thought there would be only one difference in the proof until I tried it, so I'm glad I did it, it's good practice too.

(D - a)(D - a)y = 0 implies (D - a)y = z and (D - a)z = 0.

Now (D - a)z = 0 iff z = ce^(at), so
(D - a)y = ce^(at)
Let y_p = Ate^(at) for some A and note
(D - a)(Ate^(at)) = D(Ate^(at)) -atAe^(at) = Ae^(at) + Atae^(at) - aAte^(at) = Ae^(at) = ce^(at), hence A = c so that y_p = cte^(at) solves (D - a)y = ce^(at).

Now suppose y_1 is any other solution to (D - a)y = ce^(at).
Since (D - a) is linear,
(D - a)(y_1 - y_p) = (D - a)y_1 - (D - a)y_p = cte^(at) - cte^(at) = 0 and thus w = y_1 - y_p solves (D - a)y = 0, so w = de^(at) for some d.

Again, since (D - a) is linear,
(D - a)(y_p + w) = (D - a)y_p + (D - a)w = cte^(at) + 0 = cte^(at), hence
y = y_p + w = cte^(at) + de^(at) solves (D - a)y = ce^(at) and so y also solves (D - a)(D - a)y = 0 so all solutions have the form cte^(at) + de^(at).

I think that works. Thanks for all the tips!