[proofplan]
We define the proposed solution by the displayed formula and rewrite it as $\Phi(t)$ times a varying vector. The fundamental matrix equation supplies the derivative of $\Phi$, while the [fundamental theorem of calculus](/theorems/632) for continuous vector-valued functions supplies the derivative of the integral term. These two derivatives combine so that the homogeneous part contributes $A(t)x(t)$ and the endpoint contribution of the integral contributes exactly $b(t)$. Finally, evaluating at $t=t_0$ gives the prescribed initial value.
[/proofplan]
[step:Define the candidate solution using the fundamental matrix]
Since $\Phi(t)$ is invertible for every $t\in I$, the inverse matrix $\Phi(t)^{-1}$ is defined for every $t\in I$. Define the vector $c\in \mathbb{R}^n$ by
\begin{align*}
c=\Phi(t_0)^{-1}x_0.
\end{align*}
Define the continuous map $g:I\to \mathbb{R}^n$ by
\begin{align*}
g(s)=\Phi(s)^{-1}b(s).
\end{align*}
The continuity of $g$ follows because $b$ is continuous and the inverse map is continuous on the [open set](/page/Open%20Set) of invertible matrices.
Define $H:I\to \mathbb{R}^n$ by
\begin{align*}
H(t)=\int_{t_0}^{t}g(s)\,d\mathcal{L}^1(s),
\end{align*}
with the standard oriented convention when $t<t_0$. Since $g$ is continuous, the indefinite oriented integral $H$ is continuous on $I$. Now define $x:I\to \mathbb{R}^n$ by
\begin{align*}
x(t)=\Phi(t)(c+H(t)).
\end{align*}
Substituting the definitions of $c$ and $H$ gives exactly
\begin{align*}
x(t)=\Phi(t)\Phi(t_0)^{-1}x_0+\int_{t_0}^{t}\Phi(t)\Phi(s)^{-1}b(s)\,d\mathcal{L}^1(s),
\end{align*}
because $\Phi(t)$ is independent of the integration variable $s$.
[/step]
[step:Verify the initial condition]
Since the oriented integral over the degenerate interval from $t_0$ to $t_0$ is zero, we have
\begin{align*}
H(t_0)=0.
\end{align*}
Therefore
\begin{align*}
x(t_0)=\Phi(t_0)(c+H(t_0))=\Phi(t_0)\Phi(t_0)^{-1}x_0=x_0.
\end{align*}
Thus the candidate satisfies the prescribed initial condition.
[/step]
[step:Differentiate the candidate solution]
Let $t$ be an interior point of $I$. Since $g:I\to \mathbb{R}^n$ is continuous, the fundamental theorem of calculus for continuous vector-valued functions applies to the oriented indefinite integral defining $H$. Thus $H$ is differentiable at $t$ and
\begin{align*}
\dot{H}(t)=g(t)=\Phi(t)^{-1}b(t),
\end{align*}
whether $t$ lies to the left or to the right of $t_0$.
Since $\Phi$ is a fundamental matrix, it satisfies
\begin{align*}
\dot{\Phi}(t)=A(t)\Phi(t).
\end{align*}
Using the product rule for the product of the matrix-valued map $\Phi:I\to \mathbb{R}^{n\times n}$ and the vector-valued map $c+H:I\to \mathbb{R}^n$, we obtain
\begin{align*}
\dot{x}(t)=\dot{\Phi}(t)(c+H(t))+\Phi(t)\dot{H}(t).
\end{align*}
Substituting the two derivative identities gives
\begin{align*}
\dot{x}(t)=A(t)\Phi(t)(c+H(t))+\Phi(t)\Phi(t)^{-1}b(t).
\end{align*}
Since $x(t)=\Phi(t)(c+H(t))$ and $\Phi(t)\Phi(t)^{-1}$ is the identity matrix on $\mathbb{R}^n$, this becomes
\begin{align*}
\dot{x}(t)=A(t)x(t)+b(t).
\end{align*}
[guided]
The point of introducing $H$ is to isolate the only part of the formula whose upper endpoint varies. We have defined
\begin{align*}
H(t)=\int_{t_0}^{t}\Phi(s)^{-1}b(s)\,d\mathcal{L}^1(s).
\end{align*}
The integrand is continuous because $b$ is continuous and $s\mapsto \Phi(s)^{-1}$ is continuous on $I$. The latter follows from the stated continuity of $\Phi$ on $I$, the fact that every $\Phi(s)$ is invertible, and the continuity of matrix inversion on the set of invertible matrices. Hence the fundamental theorem of calculus for continuous vector-valued functions applies to the oriented indefinite integral defining $H$. It gives that $H$ is differentiable at every interior point $t$ of $I$ and
\begin{align*}
\dot{H}(t)=\Phi(t)^{-1}b(t),
\end{align*}
with the same derivative formula on both sides of $t_0$ because the integral is oriented.
Now write the candidate solution as
\begin{align*}
x(t)=\Phi(t)(c+H(t)),
\end{align*}
where $c=\Phi(t_0)^{-1}x_0$. This form separates the homogeneous evolution, represented by $\Phi(t)$, from the accumulated forcing, represented by $H(t)$. Applying the product rule for a matrix-valued function times a vector-valued function gives
\begin{align*}
\dot{x}(t)=\dot{\Phi}(t)(c+H(t))+\Phi(t)\dot{H}(t).
\end{align*}
Because $\Phi$ is a fundamental matrix for $\dot{x}=A(t)x$, it satisfies
\begin{align*}
\dot{\Phi}(t)=A(t)\Phi(t).
\end{align*}
Substituting this identity and the formula for $\dot{H}(t)$ yields
\begin{align*}
\dot{x}(t)=A(t)\Phi(t)(c+H(t))+\Phi(t)\Phi(t)^{-1}b(t).
\end{align*}
The first term is $A(t)x(t)$ by the definition of $x(t)$, and the second term is $b(t)$ because $\Phi(t)\Phi(t)^{-1}$ is the identity matrix. Therefore
\begin{align*}
\dot{x}(t)=A(t)x(t)+b(t).
\end{align*}
This proves the differential equation at every interior point of $I$.
[/guided]
[/step]
[step:Conclude that the formula gives a global solution]
The map $x:I\to \mathbb{R}^n$ is defined for every $t\in I$ because $\Phi(t)$ is invertible for every $t\in I$ and the continuous map $g:I\to \mathbb{R}^n$ is integrable over each compact subinterval between $t_0$ and $t$. The map $H$ is continuous on $I$, and therefore $x(t)=\Phi(t)(c+H(t))$ is continuous on $I$ as a product of continuous matrix-valued and vector-valued maps. The preceding steps show that $x(t_0)=x_0$ and that $\dot{x}(t)=A(t)x(t)+b(t)$ at every interior point of $I$. Hence the displayed formula defines a solution on all of $I$ with the prescribed initial data, in the solution sense stated in the theorem.
[/step]
