Let $A = \left[ \begin{matrix} -1 & 1 \\ 0 & -1 \end{matrix} \right]$, $b = \left[ \begin{matrix} 2 \\ 3 \end{matrix} \right]$. Solve the differential equation $$\dot x = Ax + b; \quad x(0) = \left[ \begin{matrix} 1 \\ 2 \end{matrix} \right].$$
We use the formula
$$\begin{split} x(t) &= e^{tA}x(0) + \int_0^t e^{(t-s)A}b\,\text{d}s \\ &= e^{-t}\left[ \begin{matrix} 1 + 2t \\ 2 \end{matrix} \right] + \int_0^t e^{-(t-s)}\left[ \begin{matrix} 1 & t-s \\ 0 & 1 \end{matrix} \right]\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right]\text{d}s \\ &\vdots \end{split}$$
My question is how do we get from the second summand on the RHS on line 1 to the second summand on the RHS of the equation on line 2? That is how does the equation come out? $$\int_0^t e^{(t-s)A}b\,\text{d}s = \int_0^t e^{-(t-s)}\left[ \begin{matrix} 1 & t-s \\ 0 & 1 \end{matrix} \right]\left[ \begin{matrix} 2 \\ 3 \end{matrix} \right]\text{d}s$$
\begin{align} e^{(t-s)A}=e^{-(t-s)}e^{t-s}e^{(t-s)A}=e^{-(t-s)}e^{(t-s)(A+I)}=e^{-(t-s)}e^{\left[\begin{array}{cc}0&t-s\\0&0\\\end{array}\right]}=e^{-(t-s)}\left(I+\left[\begin{array}{cc}0&t-s\\0&0\\\end{array}\right]\right)=e^{-(t-s)}\left[\begin{array}{cc}1&t-s\\0&1\\\end{array}\right] \end{align} where I have used formula $e^A=I+A$, which is valid if $A^2=0$.