Proof that Lipschitz condition guarantees well posedness of initial value problems 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?
 A: It is more convenient to use the integral form: 
$$
y(x)=y_0+\int_0^x f\big(s,y(s)\big)\,ds
$$
and
$$
z(x)=y_0+\varepsilon_0+\int_0^x \big(\,f\big(s,z(s)\big)+\delta(s)\big)\,ds
$$
and hence
$$
z(x)-y(x)=\varepsilon_0+\int_0^x\delta(s)\,ds+\int_0^x\big(f\big(s,z(s)\big)-f\big(s,y(s)\big)\big)\,ds,
$$
which implies that (for $x>0$)
$$
\lvert z(x)-y(x)\rvert \le \lvert \varepsilon_0\rvert +\varepsilon x+
L\int_0^x\big|\,z(s)-y(s)\big|\,ds. \tag{1}
$$
Now, let $w(x)=\int_0^x \big|\,z(s)-y(s)\big|\,ds$, so the above becomes
$$
w'(x) \le \lvert \varepsilon_0\rvert +\varepsilon x+
L w(x)
$$
or
$$
\left(\mathrm{e}^{-Lx}w(x)\right)'=\mathrm{e}^{-Lx}\big(w'(x)-Lw(x)\big)\le 
\mathrm{e}^{-Lx}\big(\lvert\varepsilon_0\rvert +\varepsilon x\big).
$$
Integrating the above in $[0,x]$ we obtain
$$
\mathrm{e}^{-Lx}w(x)-w(0)=\frac{\lvert\varepsilon_0\rvert}{L}(1-\mathrm{e}^{-Lx})
+\varepsilon\mathrm{e}^{-Lx}\left(-\frac{x}{L}+\frac{1}{L^2}\right)-\frac{\varepsilon}{L^2}
$$
or
$$
\int_0^x\lvert z(s)-y(s)\rvert\,ds\le 
\frac{\lvert\varepsilon_0\rvert}{L}(\mathrm{e}^{Lx}-1)
+\varepsilon\left(-\frac{x}{L}+\frac{1}{L^2}\right)-\frac{\varepsilon \mathrm{e}^{Lx}}{L^2}.
$$
Using $(1)$ we obtain
$$
\lvert z(x)-y(x)\rvert \le
\lvert\varepsilon_0\rvert(\mathrm{e}^{Lx}-1)
+\varepsilon\left(-x+\frac{1}{L}\right)-\frac{\varepsilon \mathrm{e}^{Lx}}{L}+\lvert\varepsilon_0\rvert +\varepsilon x =\cdots.
$$
A: Multiply through by $|\epsilon (x )|$ to get:
$|\epsilon'(x )\epsilon (x )|\leq L|\epsilon (x )|^2+\epsilon|\epsilon (x )|$
$\frac{1}{2}|(\epsilon^2(x))'|\leq L |\epsilon^2(x)|+\epsilon|\epsilon (x)|\leq L|\epsilon(x)^2|+\epsilon^2$
$\Rightarrow \frac {1}{2}\frac {d}{dx}(|\epsilon ^ 2( x ) | e ^ { -2Lx })\leq \epsilon^2e^{-2Lx}\Rightarrow\frac {1}{2}|\epsilon ^ 2( x ) | e ^ { -2Lx}\leq \frac{1}{2}|\epsilon ^ 2 ( 0) |+\frac{1}{2L}(\epsilon ^ 2)(1-e^{-2Lx})  $
$\Rightarrow |\epsilon ^ 2( x ) |\leq|\epsilon ^ 2( 0) |e^{2Lx}+\frac{1}{L}(\epsilon^2)(e^{2Lx}-1)\leq\epsilon^2e^{2Lx}+\frac{1}{L}(\epsilon^2)(e^{2Lx}-1)=\frac{\epsilon^2}{L^2}(L^2+Le^{2Lx}-L)$
still working on this, but it's quite close to the desired estimate.
