integral, matrices, exponentiation, upper bound, time,linear system, unclear derivation How do I derive from $\dot{x}(t)=Ax(t)+Bu(t), x(0)=p$ that 
$x(t;p)=e^{At}p+\int_0^t e^{A(t-s)}Bu(s)ds$ ?
Here $A$ and $B$ are matrices of suitable sizes: $A$ is $n\times n$ and $B$ is $n \times m$,both constant real matrices, and $\dot{x}$ is derivation of $x(t)$ by $t$.
 A: Here's my derivation:
from
$\dot x(t) = Ax(t) + Bu(t) \tag{1}$
we write
$\dot x(t) - Ax(t) = Bu(t), \tag{2}$
and multiply through by $e^{-At}$:
$e^{-At}(\dot x(t) - Ax(t)) = e^{-At}Bu(t); \tag{3}$
the left-hand side may be written
$e^{-At}(\dot x(t) - Ax(t)) = e^{-At}\dot x(t) - e^{-At}Ax(t), \tag{4}$
and we note that
$\dfrac{d}{dt}(e^{-At}x(t)) =  e^{-At}\dot x(t) - Ae^{-At}x(t) = e^{-At}\dot x(t) - e^{-At}Ax(t); \tag{5}$
combining (3), (4), and (5) we find
$\dfrac{d}{dt}(e^{-At}x(t)) = e^{-At}Bu(t); \tag{6}$
we integrate (6) 'twixt $0$ and $t$, viz.
$e^{-At}x(t) - e^{-A(0)}x(0) = \displaystyle \int_0^t \dfrac{d}{ds}(e^{-As}x(s))ds = \displaystyle \int_0^t e^{-As}Bu(s)ds; \tag{7}$
we observe that 
$e^{-A(0)}x(0) = e^0x(0) = Ix(0) = x(0) = p, \tag{8}$
which we bring over to the right-hand side:
$e^{-At}x(t) = p + \displaystyle \int_0^t e^{-As}Bu(s)ds; \tag{9}$
now a final multiplication by $e^{At}$ reveals
$x(t)  = e^{At}p + e^{At}\displaystyle \int_0^t e^{-As}Bu(s)ds; \tag{10}$
so at last,
$e^{At}\displaystyle \int_0^t e^{-As}Bu(s)ds = \displaystyle \int_0^t e^{At}e^{-As}Bu(s)ds = \displaystyle \int_0^t e^{A(t - s)}Bu(s)ds; \tag{11}$
and voila:
$x(t)  = e^{At}p + \displaystyle \int_0^t e^{A(t - s)}Bu(s)ds. \tag{12}$
