Answer
True
Work Step by Step
Obtain: $Xu'(t)=b\\
u'(t)=X^{-1}b\\
\rightarrow u(t)=\int^t X^{-1}(s)b(s)ds$
The parameters $u(t)$ in the equation $x_p(t) = X(t)u(t)$,
where $X(t)$ is a fundamental matrix for $x′ = Ax$, are determined by solving the system $X(t)u′(t) = b(t)$ for $u′(t)$ and integrating the resulting vector function.
