Q:

ODE system with lower triangular coefficient matrix

Let $v_k$ satisfies the following equation
$\frac{d}{dt} v_k(t)=\sum_{i=0}^{k-1}v_k(t)$
with initial data $v_0(t=0)=1$ and $v_j(t=0)=0$.
Notice that the coefficient matrix $A$ of this ODE system (probably infinite dimensional) is a lower triangular matrix with entries one. 
I was trying to find an upper bound of $v_k(t)\leq P(t)e^{at}$ with $P$ some polynomial of $t$ (maybe with coefficient depends on $k$ or maybe not) and $a$ a constant that does not depend on $n$.
(This is equivalent to the original question I asked, but to avoid confusion I write the ODE in recursive form.)

A:

We are given the system of ODE's 
$$
\frac{dv_k}{dt} = \sum_{i=0}^{k-1} v_i(t),\qquad k \geq 1,
$$ 
with $v_0(t)$ some integrable function in $[0,\infty)$, with initial condition $v_k(0) = 0 $ for all $k\geq 1$, and $v_0(0)=1$. 
Let $||v_0||_t = \int_0^t |v_0(\tau)|d\tau$. 
Claim. The estimate
\begin{equation}
|v_k(t)| \leq ||v_0||_t + (2^k-k-1)e^t ||v_0||_t,\qquad k \geq 1
\end{equation}
holds for all $t\geq 0$. 
Proof. By induction in $k$. First note that 
$$
\frac{dv_{k+1}}{dt} = \frac{dv_{k}}{dt} + v_{k},\qquad k\geq 1.
$$
Integrating on both sides, and using the given initial conditions,
$$
v_{k+1}(t)  = v_{k}(t) + \int_0^t v_k(\tau) d \tau,\qquad k\geq 1.
$$
For $k=1$, the inequality obviously holds. For $k\geq 1$,
\begin{align}
|v_{k+1}(t)| &\leq |v_k(t)| + \int_0^t |v_k(\tau)| d\tau \\ \\
&\leq   \left(1 + (2^k-k-1)e^t \right) ||v_0||_t + \int_0^t \left[ 1 + (2^k-k-1)e^\tau  \right] ||v_0||_\tau d\tau \\
&\leq   \left( 1 + 2^k e^t -k e^t - e^t + t + (2^k-k-1)(e^t-1) \right) ||v_0||_t  \\
&\leq   \left( 1 + (2^{k+1}-(k+1)-1) e^t  \right) ||v_0||_t. 
\end{align}
Where I used $t\leq e^t-1$ from the third to last line (and some simplifications). 
For the exact solution of the system I got
\begin{equation}
v_k(t) = \int_0^t v_0(\tau) d \tau + \sum_{j=1}^{k-1} \sum_{m=1}^j \binom{j}{m} \frac{1}{m!} \int_0^t (t-\tau)^m v_0(\tau) d \tau,
\end{equation}
$k\geq 1$, $t\geq 0$. This can be obtained by the Laplace method.

