Easy way to show the uniqueness of the solution of the time dependent linear system? Consider the time dependent linear system in $\mathbb{R}^n$:
$$
\dot{x} = Ax + b(t),
$$
where $b: \mathbb{R}^n \rightarrow \mathbb{R}^n$ is continuous.
The unique solution satisfying the initial condition $x(0) = x_0$ for $t \in \mathbb{R}$ is
$$
x(t) = e^{At}x_0 + \int_0^t e^{A(t-s)}b(s)d(s), \tag{1}
$$
How to show the uniqueness in a simplest way? 
This question has been answered in a different way by I want to solve it using the following:
My try:
When we have $\dot{x}=Ax$ we can write $\frac{d}{dt}(e^{-At}x(t))=-e^{-At}Ax(t) + e^{-At}\dot{x}(t) = 0$, then $e^{-At}x(t) = c$ and letting $t = 0$, we have $c=x_0$, therefore $x(t) = e^{At}x_0$ and we are done.
Question: How can we use the same trick and show the uniqueness of $(1)$?

Suppose that there are two solutons $x,y$ such that
$$ \dot{x}=Ax+b(t),\dot{y}=Ay+b(t),x(0)=x_0,y(0)=x_0. $$
Then
$$ \frac{d}{dt}(x-y)=A(x-y) $$
from which one has
$$ x-y=e^{At}(x(0)-y(0))=0. $$
