50 Proof by checking

so again by frame 28

$$X(w)-\exp (wA)X_0={\int }_0^w\exp ((w-t)A)U(t)𝑑t.$$

$\bullet$ To check the solution works, let $t=0$ to get $X(0)=X_0$. Then we apply standard results of calculus, and work on one entry of the matrix at a time. So by the fundamental theorem of calculus, $X$ is a differentiable function of $t$, with derivative

$$\frac{d}{dt}X=A\exp (tA)X_0+e^{0}U(t)+{\int }_0^{t}A\exp (t-s)AU(s)𝑑s$$

$$=AX+U.$$