Easy way to show the uniqueness of the solution of the time dependent linear system?

Question

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 $(1)$?

Answer 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 $$ x-y=e^{At}(x(0)-y(0))=0. $$