Solving a repeated eigenvalue ODE I am trying to solve the system:
$$
\bar{x}'
=\left(\begin{array}{rr}-8 & 4 \\ 0 & -8\end{array}\right)\bar{x}+\left(\begin{array}{rr}3e^{-8t} \\ e^{-8t} \end{array}\right),\space \bar{x}(0)=\left(\begin{array}{rr}1 \\ 3 \end{array}\right)
$$
So far I am confident that with the repeated roots $-8$, we have the general solution:
$$
\bar{x}(t)=C_1\left(\begin{array}{rr}1\\ 0\end{array}\right)e^{-8t}+C_2\left(\begin{array}{rr}1\\ 0\end{array}\right)te^{-8t}+C_2\left(\begin{array}{rr}0\\ \frac{1}{4}\end{array}\right)e^{-8t}$$
Then I pick the particular solution:
$\bar{x}_p=\bar{a}e^{-8t}+\bar{b}te^{-8t}+\bar{c}t^2e^{-8t}$
$\bar{x}_p'=-8\bar{a}e^{-8t}+\bar{b}(e^{-8t}-8te^{-8t})+\bar{c}(2te^{-8t}-8t^2e^{-8t})$
Then equating: $\bar{x}_p'=A\bar{x}_p+\left(\begin{array}{rr}3e^{-8t} \\ e^{-8t} \end{array}\right)$
$e^{-8t}$:   $-8\bar{a}+\bar{b}=A\bar{a}+\left(\begin{array}{rr}3 \\ 1 \end{array}\right)$
$te^{-8t}$: $-8\bar{b}+2\bar{c}=A\bar{b}$
$t^2e^{-8t}$: $-8\bar{c}=A\bar{c}$
Solving $-8\bar{c}=A\bar{c}$:
$-8c_1+4c_2=8c_1$, so $c_2=0$
$-8c_2=-8c_2$, so $c_1=\alpha$
$\bar{c}=\left(\begin{array}{rr}\alpha \\ 0 \end{array}\right)$
Solving $-8\bar{b}+2\bar{c}=A\bar{b}$:
$-8b_1+4b_2=-8b_1+2\alpha$
$-8b_2=-8b_2+0$
$\bar{b}=\left(\begin{array}{rr}\beta \\ \frac{alpha}{2} \end{array}\right)$
But I am no where close to finding $\bar{a}$, $\bar{b}$ and $\bar{c}$ to finish off my solution as I have too many free variables. Can anyone see where I am going wrong as I am slightly confused? Or offer a better slicker method as mine seems quite clunky...
 A: We are given
$$x' = Ax + g = 
\left(\begin{array}{rr}-8 & 4 \\ 0 & -8\end{array}\right)x+\left(\begin{array}{rr}3e^{-8t} \\ e^{-8t} \end{array}\right),\space x(0)=\left(\begin{array}{rr}1 \\ 3 \end{array}\right)$$
We find eigenvalues / eigenvectors and have a complementary solution
$$ x_c(t)=e^{-8t}\left(c_1\begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_2\left(\begin{pmatrix} 1 \\ 0 \end{pmatrix}t + \begin{pmatrix} 0 \\ \dfrac{1}{4} \end{pmatrix} \right)\right)$$
Because of the eigenvalues and the non-homogenous terms, we choose
$$x_p(t) = e^{-8t}(\vec a + \vec b t + \vec ct^2)$$
This gives
$$x_p'(t) = e^{-8t}((-8 \vec a + \vec b) + (-8\vec b + 2 \vec c) t + (-8\vec c)t^2)$$
We have
$$x_p'(t) = A x_p(t) + \vec g \implies \vec g = x_p'(t) - A x_p(t)$$
Writing this out (the exponential terms divide out)
$$\begin{pmatrix} 3 \\ 1 \end{pmatrix} = \begin{pmatrix} b_1-4a_2 \\ b_2 \end{pmatrix}+ \begin{pmatrix} 2 c_1 - 4 b_2 \\ -2c_2 \end{pmatrix}t + \begin{pmatrix} -4c_2 \\ 0 \end{pmatrix}t^2$$
From this, we get
$$a_1 = a_2 = c_2 = 0, b_1 = 3, b_2 = 1, c_1 = 2$$
We can now write
$x(t) = x_c(t) + x_p(t) =e^{-8t}\left(c_1\begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_2\left(\begin{pmatrix} 1 \\ 0 \end{pmatrix}t + \begin{pmatrix} 0 \\ \dfrac{1}{4} \end{pmatrix} \right)\right) + e^{-8t}\begin{pmatrix} 2t^2+ 3t \\ t \end{pmatrix}$
Using this $x(t)$ with the IC, we get
$$x(t) = e^{-8t}\begin{pmatrix} 2t^2+15t+1 \\ t+3 \end{pmatrix}$$
Note: We used Undetermined Coefficients to solve this system and that is only one approach, there are many other methods.
