Solving a linear system using matrix exponentials

Question

Let $A$ be a certain $6\times 6$ matrix. Find the solution to the initial value problem $\mathbf{x}'=A\mathbf{x}$; $\mathbf{x}(2)=\begin{bmatrix}1 \\ 3 \\ 5 \\ 7 \\ 9 \\ 11\end{bmatrix}$.

My attempt:

I've managed to do most of the work this problem, which is calculating $e^{At}$.

The trouble is that $\mathbf{x}(2)$ is given rather than $\mathbf{x}(0)$.

The general solution is given by $\mathbf{x}(t)=e^{At}\mathbf{x}(0)$. I could plug in $t=2$ to get $\mathbf{x}(2)=e^{2A}\mathbf{x}(0)$ and then find $\mathbf{x}(0)$ but I strongly suspect there's a simpler way to do this.

Any suggestions?

Answer 1

The general solution is $\bf x(t)=\exp(\bf A t) \bf c =\exp(\bf A t) \exp(-\bf A t_0) \bf c'= \exp(A(t-t_0) \bf c'$.

Then $t_0=2 \quad x(t_0)=x(2)=$as given, ...

Answer 2

$\mathbf{x}(0)=e^{-2A}\mathbf{x}(2)$

$\implies\mathbf{x}(t)=e^{At}e^{-2A}\mathbf{x}(2)=e^{A(t-2)}\mathbf{x}(2)$