$\displaystyle
\frac{dx}{dt} = a(t)x + f(t), \qquad x(0) = x_0. $
(1)
Let
\[
\mu(t)= e^{\int_0^t a(s) ds}.
\]
Theorem A.
If the functions $a(t)$ and $f(t)$ are continuous, then the initial value
problem
(1) has exactly one solution, which is given by
\begin{equation}
x(t) = \mu(t) \Bigl\{x_0 + \int_0^t \frac{f(s)}{\mu(s)} ds\Bigr\}.
\label{eq:solution}
\end{equation}