Laplace transform for IVP not at zero in system of differential equations

Question

Suppose we have a system $\boldsymbol X'=\boldsymbol A\boldsymbol X$. Let's denote the laplace transform of a vector $\boldsymbol Y$ as $\mathscr L\{\boldsymbol Y(t)\}(s)=\boldsymbol y(s)$. If we apply the laplace transform to our system: $$s\boldsymbol x-\boldsymbol X(0)=\boldsymbol A \boldsymbol x\Longrightarrow \boldsymbol x=(s\boldsymbol I-\boldsymbol A)^{-1}\boldsymbol X(0).$$

The solution is obtained when we apply the inverse laplace transform. How could I solve such a system when I'm given a vector not at $0$? I haven't come across with an exercise like that one yet, but I wonder how could one solve it...

Answer 1

To solve $X' = A X$ with $X(t_0) = b$ given, solve $Y' = A Y$ with $Y(0) = b$, and take $X(t) = Y(t - t_0)$.

Answer 2

I edited Robert's answer. However it was not accepted. Here there is a more explicit explanation of his idea:

Consider the change of variables:

$$\boldsymbol X(t)=\boldsymbol Y(t-t_0).$$

Therefore the system

$$\cases{\boldsymbol X'(t)=\boldsymbol A\boldsymbol X(t)\\\boldsymbol X(t_0)= \boldsymbol b}$$ is equivalent to the system

$$\cases{\boldsymbol Y'(t-t_0)=\boldsymbol A \boldsymbol Y(t-t_0)\\ \boldsymbol Y(0) = \boldsymbol b}.$$

If you solve the last system you will obtain a solution $\boldsymbol Y(t-t_0)=\boldsymbol X(t)$ which is the solution to the initial system.