In order to prove that solutions of linear equations with constant coefficients grow at most exponentially, you can use a multidimensional version of Gronwall's lemma.

Remember indeed that the solution $\displaystyle y(t)$ of a $\displaystyle n$-th order linear differential equation is a component of the solution $\displaystyle Y(t)$ of a $\displaystyle n$-dimensional linear equation. Then an exponential bound on the euclidean norm of $\displaystyle Y(t)$ gives an exponential bound on $\displaystyle y(t)$. Here's how to obtain such a bound.

If $\displaystyle Y'(t)=AY(t)+B$ where $\displaystyle A,B$ are $\displaystyle n\times n$-matrices, we have $\displaystyle \|Y'(t)\|\leq\|A\|\|Y(t)\|+\|B\|$ for every $\displaystyle t\geq 0$, so that $\displaystyle \|Y(t)-Y(0)\|\leq \|A\| \int_0^t \|Y(s)\| ds + \|B\| t$, hence $\displaystyle \|Y(t)\|\leq \|Y(0)\|+\|B\|t+\|A\|\int_0^t \|Y(s)\| ds$. And now you can apply the one-dimensional Gronwall lemma "in integral form" (quoting the

wikipedia) to the function $\displaystyle t\mapsto\|Y(t)\|$. The only online reference I could find for the final result is this french

exercise sheet (Exercice 1.4), but you don't need it to find the proof, just apply Gronwall's lemma like I wrote and do an integration by parts to compute the integral. You should get:

$\displaystyle \|Y(t)\|\leq \|Y(0)\|e^{\|A\|t}+\frac{\|B\|}{\|A\|}(e^{\|A\|t}-1)$.

This is what you need.