Answer
False
Work Step by Step
If $x_0(t)$ is a solution to the homogeneous vector differential equation $x'(t)=A(t)x(t)$, let's substitute $x(t)=x_0(t)+b(t)$ into $x'(t) $ we have:
$$x'(t)=A(t)[x_0(t)+b'(t)]\\
=A(t)x_0(t)+A(t)b'(t)$$
which is not same as given in the statement.
