1. ## Ode

Find a general sol'n for the following ODE:

$\displaystyle y''-6y'+9y=2+te^{4t}$

So I use the auxiliary eq:

m^2 - 6m + 9 = 0

m = 3.

So, the homogeneous sol'n, y_c, is:

$\displaystyle y_c = c_1e^{3t}$

How'd I complete it?

2. Good job so far.

What needs to be done next is find the parts of the solution that are linearly independent to $\displaystyle e^{3t}$. This is done by substituting y=$\displaystyle ue^{3t}$. This will net you both the 2nd part of the homogeneous solution and the particular solution.

y =$\displaystyle ue^{3t}$
y' = $\displaystyle u'e^{3t}+3ue^{3t}$
y'' = $\displaystyle u''e^{3t}+6u'e^{3t}+9ue^{3t}$

So our equation becomes:

$\displaystyle (u''e^{3t}+6u'e^{3t}+9ue^{3t})-6(u'e^{3t}+3ue^{3t})+9ue^{3t}$ = $\displaystyle 2+te^{4t}$
$\displaystyle u''e^{3t}$ = $\displaystyle 2+te^{4t}$

notice that the u term has cancelled out. This will always happen when we make this substitution. In this case the u' term has also been cancelled but this was just lucky.

u'' = $\displaystyle 2e^{-3t}+te^t$
u' = $\displaystyle \int 2e^{-3t}+te^t$dt
=$\displaystyle -\frac {2}{3}e^{-3t}+te^t-e^t+c_2$(I have integrated by parts)
u = $\displaystyle \int -\frac {2}{3}e^{-3t}+te^t-e^t+c_2 dt$
= $\displaystyle \frac {2}{9}e^{-3t}+te^t-e^t-e^t+c_2t+c_1$
The reason for reusing $\displaystyle c_1$ will soon be apparent
u =$\displaystyle \frac {2}{9}e^{-3t}+te^t-2e^t+c_2t+c_1$

y = $\displaystyle ue^{3t}$
= $\displaystyle (\frac {2}{9}e^{-3t}+te^t-2e^t+c_2t+c_1)e^{3t}$
=$\displaystyle \frac {2}{9}+te^{4t}-2e^{4t}+c_2te^{3t}+c_1e^{3t}$