In the proof of the theorem which states that the Lipschitz condition guarantees well posed-ness of an initial value problem $y'=f(x,y)$, $y(x\_0)=y\_0$, I came across this

Let the perturbed problem be

$$
z'=f(x,z) + \delta(t) \\
z(x\_0)=y\_0 + \epsilon\_0
$$ 

Let $\epsilon(x)=z(x)-y(x)$ and let $f$ satisfy the Lipschitz condition. Then we have

$$
\epsilon '=z'-y'=f(x,z)-f(x,y)+\delta(x)\\
\implies |\epsilon'(x)| \leq |f(x,z)-f(x,y)| + |\delta(x)|\\
$$

Let $\delta(x)$ have the maximum value $\epsilon$. Therefore, we get,

$$
|\epsilon'(x)| \leq |f(x,z)-f(x,y)| + \epsilon\\
\implies |\epsilon'(x)| \leq L|\epsilon(x)| + \epsilon
$$

I don't understand the next step i.e. how is the above inequality integrated to give

$$
|\epsilon(x)| \leq \frac{\epsilon}{L} \left[ (L+1)e^{Lt}-1\right] 
$$

How is the inequality integrated?