Could anyone show me how to calculate the general solution to $$x' = \begin{pmatrix} 3 & -1 \\ 4 & -1 \end{pmatrix}x+\begin{pmatrix} 1 \\ t \end{pmatrix}$$ where $x_1(0) = 1$ and $x_2(0) = 0$?
I get the fundamental matrix $$e^{tA} = e^t+(A-I_2)te^{t}$$
And then I want to use a theorem that says that the general solution will be on the form $$x(t) = F(t)c+F(t) \int_{t_0}^{t} F^{-1}(\tau)b(\tau)d\tau$$ where $F(t)$ is the fundamental matrix, and c is a column-matrix dependent on the initial value given.
In this case, this would be: $$x(t) = e^{tA}\begin{pmatrix} 1 \\ 0 \end{pmatrix}+e^{tA} \int_{0}^{t} e^{-\tau A}\begin{pmatrix} 1 \\ \tau \end{pmatrix} d\tau$$
But then it feels like the computations get nasty, when you substitute $$e^t+(A-I_2)te^t$$ for $e^{tA}$. Is there anyway to simplify this?
For the Fundamental Matrix, we get
$$e^{tA} = e^t + (A-I_2)te^{t} = e^t \left(I + t \begin{pmatrix} 2 & -1 \\ 4 & -2 \\ \end{pmatrix} \right) = e^t \begin{pmatrix} 2 t+1 & - t \\ 4 t & -2t+1 \\ \end{pmatrix}$$
Can you now continue?
Update
The intermediate calculations are
$e^{-\tau A}\begin{pmatrix} 1 \\ \tau \end{pmatrix} = \begin{pmatrix} e^{-\tau} \tau^2+e^{-\tau} (1-2 \tau) \\ e^{-\tau} \tau (2 \tau+1)-4 e^{-\tau} \tau \\ \end{pmatrix}$
$\displaystyle \int_0^t e^{-\tau A}\begin{pmatrix} 1 \\ \tau \end{pmatrix}d\tau = \begin{pmatrix} -e^{-t} t^2-e^{-t}+1 \\ -2 e^{-t} t^2-e^{-t} t-e^{-t}+1 \\ \end{pmatrix}$
$\displaystyle e^{tA} \int_0^t e^{-\tau A}\begin{pmatrix} 1 \\ \tau \end{pmatrix}d\tau = \begin{pmatrix} e^t t-t+e^t-1 \\ 2 e^t t-3 t+e^t-1 \\ \end{pmatrix}$
The final solution is
$$X(t) = \begin{pmatrix} x_1(t)\\ x_2(t) \end{pmatrix} = e^{tA}\begin{pmatrix} 1 \\ 0 \\ \end{pmatrix} + e^{tA} \int_0^t e^{-\tau A}\begin{pmatrix} 1 \\ \tau \end{pmatrix}d\tau = \begin{pmatrix} e^t (3 t+2)-t-1 \\ e^t (6 t+1)-3t -1 \\ \end{pmatrix}$$