In Laplace space:
$$ sX(s)-x(0)=AX(s)+BU(s) $$
so:
$$ X(s)=\frac{BU(s)}{sI-A}+\frac{x(0)}{sI-A} $$$$ X(s)=(sI-A)^{-1}BU(s)+(sI-A)^{-1}x(0) $$
the second term, due to the initial condition, produces:
$$ x_0(t)=x(0)e^{At} $$$$ x_0(t)=e^{At}~x(0) $$
while the fist term, due to the incoming signal, produces:
$$ x_u(t)=\int_0^tBe^{At}u(t-\tau)dτ=\int_0^tBe^{A(t-\tau)}u(t)dτ $$$$ x_u(t)=\int_0^te^{A\tau}~B~u(t-\tau)d\tau=\int_0^te^{A(t-\tau)}~~B~u(\tau)~d\tau $$
(where calculation of \$ e^{At} \$ can be seen here )
Finally:
$$ y(t)=Cx(t)=C(x_u(t)+x_0(t)) $$