I'm trying to derive this solution given to us in a systems / control-theory course without proof.
The first-order vector linear differential equation in question is
$$\dot{x}(t)-Ax(t) = f(t)$$ $$x(0) = x_0$$
Where $x$ is a $n$-dimensional vector in $R^n$ (for instance, a two or three dimensional vector), and $A$ is an $n$ x $n$ constant matrix.
The unexplained solution is $$x(t) = e^{At}x_0 + \int_0^t e^{A(t-\tau)}f(\tau)d\tau$$
Where the first term $e^{At}x_0$ is the homogenous solution to
$$\dot{x}(t)-Ax(t) = 0$$ $$x(0) = x_0$$
and the second term is the "particular" solution, the one I'm interested in deriving. I think it comes from a Green's function approach where $f(t)$ is decomposed into an infinite sum of dirac delta functions $\delta$:
$$ f(t) = \int_0^\infty f(\tau)\delta(t-\tau)d\tau $$
So by linearity of the differential equation, we first find a solution (response, trajectory, aka Green's function, $G(\tau,t)$) for a single delta function "system input" at time $\tau$ and add them up invoking the linearity of the differential equation to obtain: $$ x(t) = \int_0^\infty G(t, \tau)f(\tau)d\tau $$
But I'm a bit confused how to set up the ODE to get the Green's function. Following wiki and other sources, I have: $$ \dot{G}(t,\tau) - A G(t, \tau) = \delta(t-\tau) $$ but this $\delta$ has to be vector-valued since the left side is. And I suppose this "dirac vector" should point parallel to $f(\tau)$: let $\hat{f}(\tau)$ be this unit vector to get it pointing properly. $$ \dot{G}(t,\tau) - A G(t, \tau) = \hat{f}(\tau) \delta(t-\tau) $$
Then there is the question of what initial value to use with this ODE. I don't really know how to answer that and why. But I tried $G(t=0,\tau)=0$ for all $\tau$. I figured if the initial condition wasn't zero it would mess up the homogenous solution obeying a non-zero initial condition(?).
Then the ODE says the system does nothing until $t > \tau$, after which it does its natural (homogenous) response after a sudden jump from the origin to the point $\hat{f}(\tau)$: $$ G(t,\tau) = 0\space for \space 0<t<\tau $$ $$ G(t,\tau) = e^{A(t-\tau)} \hat{f}(\tau) \space for \space 0<\tau<t $$ Which can be captured in one line by using the step function $H(t-\tau)$ ($1$ if $t>\tau$, $0$ otherwise): $$ G(t,\tau) = H(t-\tau) e^{A(t-\tau)} $$
Then adding up all the responses (ODE is linear, more justification desired) gives
$$ x(t) = \int_0^\infty H(t-\tau) e^{A(t-\tau)} f(\tau) d\tau $$
The integrand is zero if $\tau$ exceeds t, so it simplifies to
$$ x(t) = \int_0^t e^{A(t-\tau)} f(\tau) d\tau $$
All of which tries to explain the particular solution.
Does this argument seem ok? Any areas where you could better explain what is going on? Specifically, I'm not sure if my reasoning about the boundary condition / initial value on $G$ is correct. Also my justification or reasoning for using $\hat{f}(\tau)$ to give my dirac delta function a vector value seems a little hand-wavy. Thanks.
You get that much faster by considering the exponential as integrating factor. Then $$ \frac{d}{dt}(e^{-At}x(t))=e^{-At}(\dot x(t)-Ax(t))=e^{-At}f(t) $$ and the claim follows from the fundamental theorem of calculus. $$ e^{-At}x(t)-e^{-A0}x(0)=\int_0^te^{-Aτ}f(τ)dτ \\ x(t)=e^{At}x_0+\int_0^te^{A(t-τ)}f(τ)dτ $$
As you found out in the end, $G$ is matrix-valued. Thus the right side also has to be matrix valued from the start, $δ(t−τ)I$ with the identity matrix. Then the other parts of that calculation also fit together.