3rd Order Differential Equation Help

Find the general solution of...

y''' - 3y'' + 2y' = t + e^t

The main part I'm having trouble with is doing the method of undetermined coefficients for the right hand side of the equation. Any help is appreciated.

The equation y''' - 3y'' + 2y' = t + e^t
can be transformed into y'' - 3y' + 2y = t + e^t by replacing y' with y.

First, we solve the equation y'' - 3y' + 2y = 0.
Let y = e^mt, so that y' = me^mt and y'' = m^2 e^mt.
Substituting these back into y'' - 3y' + 2y = 0, we obtain:
m^2 e^mt - 3(me^mt) + 2(e^mt) = 0
e^mt (m^2 - 3m + 2) = 0
e^mt ≠ 0 for any t so m^2 - 3m + 2 = 0.
Therefore (m - 2)(m - 1) = 0,
so m = 2, m = 1 are solutions.
So the complementary function is y = Ae^2t + Be^t

Now we need to deal with y'' - 3y' + 2y = t + e^t.
Since there's a t on the right-hand side of the equation, we need what we substitute in to have (a + bt) in it. Since there's an e^t, we would usually have a (ce^t) term, but, since there's already an e^t term in our complementary function, we need to multiply this by t to make (cte^t). Putting this all together, we find we need to substitute in:
y = a + bt + cte^t.
Therefore y' = b + ce^t + cte^t, and
y'' = ce^t + ce^t + cte^t = 2ce^t + cte^t
Substituting these back into y'' - 3y' + 2y = t + e^t, we obtain:
(2ce^t + cte^t) - 3(b + ce^t + cte^t) + 2(a + bt + cte^t) = t + e^t
2ce^t + cte^t - 3b - 3ce^t -3cte^t + 2a + 2bt + 2cte^t = t + e^t
(2a - 3b) + (2b)t + (-c)e^t = t + e^t
Comparing coefficients, we get:
2a - 3b = 0 (Equation 1),
2b = 1 (Equation 2), and
-c = 1 (Equation 3).
Rearranging Equations 2 and 3, we get c = -1 and b = 1/2.
By substituting b = 1/2 back into Equation 1, we get:
2a - 3(1/2) = 0