This is a passage from Ferdinand verhult's ``Nonlinear Differential Equations and Dynamical Systems'' (Theorem 6.2):
Consider the differential equation $$ \dot{x}=Ax+B(t)x, $$ $A$ $n\times n$ real matrix, $B(t)$ real continuous, defined for $t\geq t_0$.
Let $\Phi(x)$ be the fundamental matrix of $\dot{x}=Ax$, which can be written as $$ \Phi(t)=\exp(A(t-t_0)). $$
Substituting $x=\Phi(t)z$ in the ODE yields
$$ \dot{z}=\Phi^{-1}(t) B(t) \Phi(t). $$
The author then says "Integration of this expression and multiplication with $\Phi(t)$ produces for the solutions of the ODE the integral equation:"
$$ x(t)=\Phi(t) x_0 +\int_{t_0}^t \Phi(t-\tau+t_0)B(\tau)x(\tau) d\tau $$
Here it's used the fact that $\Phi(t)\Phi^{-1}(\tau)=\Phi(t-\tau+t_0)$.
I can't seem to understand how it is being integrated here. Would someone care to explain? I seem to be missing something really simple here.
I know this is from variation of constants and that there are some other formulations for the expression of the solutions but I really need the integration step in here. Thanks in advance.
There are some typo: the substitution should be $x=\Phi(t)z$, and we have $$ \Phi(t)\Phi^{-1}(\tau)=\Phi(t-\tau+t_0). $$ From the equation for $z$ after the substitution, $$ \dot{z}=\Phi^{-1}(t)B\Phi(t)z, $$ We have $$ \dot{z}=\Phi^{-1}(t)B x(t). $$
Replace $t$ by $\tau$ and integrate from $t_0$ to $t$ gives $$ z(t)=z(t_0)+\int_{t_0}^t \Phi^{-1}(\tau)B x(\tau) d\tau. $$ Multiply $\Phi(t)$ on the left, then $$ x(t)=x(t_0)+\Phi(t)\int_{t_0}^t \Phi^{-1}(\tau)B x(\tau) d\tau. $$ This is the result.