Consider $$\dot{Y}=(A+\epsilon B)Y,\quad Y(0)=I ,$$ where $A, B\in\mathbb{R}^{n \times n}$. Prove that the matrix solution $Y$satisfies the integral equation $$ Y(t)=X(t)+\epsilon\int_0^tX(t-s)BY(s)\,\mathrm{d}s, $$ where $X$ is the solution of the IVP $$ \dot{X}=AX, \quad X(0)=I. $$
I just learned the first-order linear equations. It is easy to see that $$ Y(t)=e^{t(A+\epsilon B)}, \ X(t)=e^{tA}. $$ Substitute into the integral equation, if $[A,B]=AB-BA=0$, the conclusion is trivial. However, I do not know how to prove the general case. Maybe the $\epsilon$ is vital, but I cannot figure out the details.
Appreciate any help!
Let $sI-A=P$. Then \begin{align}Y(t)-X(t)&=\mathcal L^{-1}[(sI-A-\epsilon B)^{-1}]-\mathcal L^{-1}[(sI-A)^{-1}]\\&=\mathcal L^{-1}[(P-\epsilon B)^{-1}-P^{-1}]\end{align} and convolution gives \begin{align}\epsilon\int_0^tX(t-s)BY(s)\,ds&=\epsilon\int_0^te^{(t-s)A}Be^{s(A+\epsilon B)}\,ds\\&=\epsilon\mathcal L^{-1}[\mathcal L[e^{tA}]\mathcal L[Be^{t(A+\epsilon B)}]]=\mathcal L^{-1}[P^{-1}\epsilon B(P-\epsilon B)^{-1}].\end{align} Hence it suffices to show that $$(P-\epsilon B)^{-1}-P^{-1}=P^{-1}\epsilon B(P-\epsilon B)^{-1}$$ which is true on left-multiplying by $P$ and right-multiplying by $P-\epsilon B$.