Differential equation involving exponential of matrix I need to prove that the initial value problem $x'(t) = Ax(t)+b(t), \,x(0) = x_0 \in \mathbb{R^n}$ can be solved by the following solution $$x(t) = \exp(At)x_0 + \int_0^t\exp((t-s)A)b(s)ds,$$ where $A: \mathbb{R} \rightarrow \mathbb{R^{n\times n}}, b: \mathbb{R} \rightarrow \mathbb{R^n}$ are differentiable and $b$ is bounded. I found that $x'(t) = A\exp (At)x_0 + b(t) - \exp(At)b(0) .$ At the other hand, we get for the right hand side of the differnetial equation, $$Ax(t) + b(t) = A\exp (At)x_0 + b(t) + A\int_0^t\exp((t-s)A)b(s)ds.$$ Thus, the proposed solution will satisfy the initial value problem, if $$A\int_0^t\exp((t-s)A)b(s)ds = - \exp(At)b(0)$$holds. I do not know how to prove this. Do you have any suggestion ? Thanks.
 A: Consider the new vector
$\mathbf{z}(t)
= \exp[-t\mathbf{A}]\mathbf{x}(t)$.
It holds using product rule
\begin{eqnarray*}
\dot{\mathbf{z}}(t)
&=&
\exp[-t\mathbf{A}] \dot{\mathbf{x}}(t)
-\mathbf{A}\exp[-t\mathbf{A}]\mathbf{x}(t) \\
&=&
\exp[-t\mathbf{A}] 
\left[
\mathbf{A}\mathbf{x}(t)+\mathbf{b}(t)
\right]
-\exp[-t\mathbf{A}] \mathbf{A}\mathbf{x}(t) \\
&=&
\exp[-t\mathbf{A}] 
\mathbf{b}(t)
\end{eqnarray*}
Integrating yields the requested form
$$
\mathbf{z}(t)-\mathbf{z}(0)
=
\int_0^t
\exp[-s\mathbf{A}] 
\mathbf{b}(s)
ds
$$
Premultiplying (by the left) by
$\exp[t\mathbf{A}]$ yields
\begin{eqnarray*}
\mathbf{x}(t)
&=&
\exp[t\mathbf{A}]\mathbf{x}(0)
+
\exp[t\mathbf{A}]
\int_0^t
\exp[-s\mathbf{A}] 
\mathbf{b}(s)
ds \\
&=&
\exp[t\mathbf{A}]\mathbf{x}(0)
+
\int_0^t
\exp[(t-s)\mathbf{A}] 
\mathbf{b}(s)
ds
\end{eqnarray*}
A: By linearity (the "superposition principle"), we can assume that $x_0 = 0$, so we want to solve
$$x'(t) - Ax(t) = b(t), x(0) = 0.$$
We already know that the solution to the homogenous problem
$$x'(t) - Ax(t) = 0, x(0) = x_0$$
is $x(t) = e^{tA}x_0$. Now for each $s \in \mathbb{R}$, let $x(t; s)$ be the solution to
$$x'(t; s) - Ax(t; s) = 0, x(s; s) = b(s).$$
Set
$$x(t) = \int_{0}^{t}x(t; s)\,ds.$$
This is exactly the formula you have for $x(t)$. We have $x(0) = 0$ as required. To compute $x'(t)$, write $g(t_1, t_2) = \int_{0}^{t_1}x(t_2; s)\,ds$. Note that $x(t) = g(t, t)$. By the chain rule, fundamental theorem of calculus and differentiation under the integral sign,
$$x'(t) = g_{t_1}(t, t) + g_{t_2}(t, t) = x(t, t) + \int_{0}^{t}x'(t; s)\,ds = b(t) + Ax(t).$$
The above method of going from the homogenous solution to the imhomogenous solution is called Duhamel's principle.
