Gronwall's type inequality with $\Vert A(t)-B(t)\Vert\le \varepsilon$ 
I have two solutions $u,v$ of $u'=Au$ and $v'=Bv$ where $A,B$ are two continuous linear maps from an interval $J\to\mathcal{L}(\Bbb{R}^n)$ such that there exists $\varepsilon$ such that $\Vert A(t)-B(t)\Vert\le \varepsilon$ for all $t\in J$ and $u(0)=v(0)$
I would like to prove that for all $[a,b]\subset J$ we have: $$\Vert u(t)-v(t)\Vert\le \frac{C\varepsilon}{K}(\exp(K\vert t\vert-1)$$ for all $t\in [a,b]$ and $C,K$ are constant to find.

So I write $$\Vert u'(t)-v'(t)\Vert=\Vert \int_0^t A(s)u(s)-B(s)v(s)ds\Vert.$$
Now I just need to prove that the RHS is less than $\int_0^t\Vert (\varepsilon+B(s))(u(s)-v(s)ds\Vert$ because then I can use the same idea of the proof of Gronwall's inequality.
But I am stuck, because I have $\Vert A(t)\Vert\le\Vert B(t)\Vert+\varepsilon$, but I don't have what I 'want'.
 A: This is not a solution, but it's too long be to a comment. 
Implicitly writing the solutions as
\begin{align}
u(t) = u(0) + \int^t_0 A(s)u(s)\ ds
\end{align}
and
\begin{align}
v(t) = v(0) + \int^t_0 B(s)v(s)\ ds
\end{align}
then we have
\begin{align}
\|u(t)-v(t)\| =&\ \Big\|\int^t_0 A(s)u(s)-B(s)v(s)\ ds \Big\|\\
\leq&\ \int^t_0\|A(s)u(s)-B(s)v(s)\|\ ds\\
=&\ \int^t_0 \|A(s)u(s)-A(s)v(s)+A(s)v(s)-B(s)v(s)\|\ ds\\
\leq&\ \int^t_0 \|A(s)\|\|u(s)-v(s)\| + \|A(s)-B(s)\|\|v(s)\|\ ds\\
\leq&\ \int^t_0\|A(s)\|\|u(s)-v(s)\| + \epsilon\|v(s)\|\ ds
\end{align}
when $t \in J$. 
Moreover, we know that
\begin{align}
v(t) = \exp\left(\int_0^t B(\tau)\ d\tau\right)v_0
\end{align}
which means
\begin{align}
\|v(t)\| \leq \exp\left(\int^t_0 \|B(\tau)\|\ d\tau \right)\|v_0\|. 
\end{align}
Thus, it follows 
\begin{align}
\|u(t)-v(t)\| \leq \int^t_0 \|A(s)\|\|u(s)-v(s)\|\ ds + \epsilon \int^t_0\exp\left(\int^s_0 \|B(\tau)\|\ d\tau \right)\|v_0\|\ ds.
\end{align}
Without further information on $A(t)$ and $J$, there's really nothing I can do.  
Edit: Consider the following example
\begin{align}
x' = \exp(t) x \ \ \text{ and } \ \ y'=(\exp(t)+\epsilon) y
\end{align}
with initial condition say $x(0) = y(0) = e^e$. Then we see the solutions are given by
\begin{align}
x(t) = \exp(\exp(t)) \ \ \text{ and } \ \ y(t) = \exp(\exp(t)+\epsilon t)
\end{align}
which means
\begin{align}
|x(t) -y(t)| = \exp(\exp(t))|\exp(\epsilon t)-1|.
\end{align}
Hence your above claim is false if you did not restrict to a subinterval $[a, b]$. 
A: This is not a solution, but is too long for a comment.
The question is a bit vague. If the question is can I find $C,K$ such that the bound holds for a fixed $A,B$ satisfying the $\epsilon$ criterion, then
we can always find a $C,K$ (that depend on $A,B$) for which the bound is
true. But this is saying nothing more than I can divide by a non zero scalar.
If we are looking for bounds that are valid for all $A,B$ then such bounds
do not exist.
To see this, consider the systems $\dot{u} = Au$, $\dot{v} = Bv$ where $A,B$ are scalars.
The solutions are $u(t) = u(0) e^{At}, v(t) = u(0) e^{Bt}$ and we see that
$|u(t)-v(t)| = |u(0)| e^{Bt}| e^{(A-B)t} -1 |$.
So, even if $|A-B|$ is bounded (and non zero) and $t\neq 0$, we can choose $B$ such that $|u(t)-v(t)|$is unbounded.
Perhaps (i) $A$ is fixed and $B$ is a perturbation?
Or perhaps (ii) $A,B$ lie in some bounded set?
In both of the latter cases, we can find $C,K$ such that the
inequality holds (in fact it could be bounded by $\|A-B\|_\infty$
rather than $\epsilon$, which is a little stronger than uniformity).
Here is an approach that shows how to address (i): Let
$\delta = u-v$, then we have
$\dot{\delta} = A \delta + (A-B)v$. Let $\phi$ be the state transition function for $\dot{u} = Au$, then we have
$\delta(t) = \int_0^t A(\tau) \phi(t,\tau) (A(\tau)-B(\tau)) v(\tau)d \tau $, from which we can obtain a useful bound.
