Inequality with differential equations solutions I would love some help working through this problem:
Let $f_1,f_2,f : [0,\infty) \to \mathbb{R}$ be three bounded, continuous and absolutely Riemann integrable functions so that $|f_1(x)|, |f_2(x)| \leq |f(x)|$, for every $x \in [0,\infty)$.  Given the system of differential equations:$$\alpha'(x)=f_1(x)\beta(x)\quad and\quad \beta'(x)=f_2(x)\alpha(x)\quad with\quad \alpha(0)=a, \beta(0) = b
$$ show that solutions $\alpha$ and $\beta$ can be expressed as infinite series:$$\alpha(x)=a+b\int_0^xf_1(y)dy+a\int_0^xf_1(y)\int_0^yf_2(z)dzdy+b\int_0^xf_1(y)\int_0^yf_2(z)\int_0^zf_1(w)dwdzdy..$$ $$\beta(x)=b+a\int_0^xf_2(y)dy+b\int_0^xf_2(y)\int_0^yf_1(z)dzdy+a\int_0^xf_2(y)\int_0^yf_1(z)\int_0^zf_2(w)dwdzdy..$$
and show that as a consequence of this, both $\alpha$ and $\beta$ are bounded functions and moreover they satisfy the estimate$$|\alpha(x)|,|\beta(x)| \leq max(|a|,|b|)exp(||f||_1)$$
for every $x \in [0,\infty)$, where $||f||_1 = \int_0^\infty|f(y)|dy$.
I have already shown that $\alpha$ and $\beta$ are solutions but I am not sure how to go about showing that they are bounded and that the inequality holds.  The textbook that I am using suggests that this can be shown by using Picard iteration but I'm not sure how to do this.
I'd appreciate any help with this.
Thanks!
 A: The system is 
$$
\begin{bmatrix}
α'(x)\\β'(x)
\end{bmatrix}
=
\begin{bmatrix}
0&f_1(x)\\f_2(x)&0
\end{bmatrix}
·
\begin{bmatrix}
α(x)\\β(x)
\end{bmatrix}
$$
or as Picard integral equation
$$
\begin{bmatrix}
α(x)\\β(x)
\end{bmatrix}
=
\begin{bmatrix}
a\\b
\end{bmatrix}
+
\int_0^x
\begin{bmatrix}
0&f_1(y)\\f_2(y)&0
\end{bmatrix}
·
\begin{bmatrix}
α(y)\\β(y)
\end{bmatrix}
dy
$$
which gives the time- or path-ordered exponential solution formula.

As a linear ODE with continuous coefficient matrix the solution exists everywhere and is unique per Picard-Lindelöf, a Lipschitz constant over the interval $[-N,N]$ is the maximum of $|f(x)|$ over that interval.
Applying the max norm to the integral equation gives 
$$
\max(|α(x)|,|β(x)|)\le \max(|a|,|b|)+\int_0^x\max(|f_1(y)|,|f_2(y)|)·\max(|α(y)|,|β(y)|)\,dy
$$
Name $e(x)=\max(|α(x)|,|β(x)|)$, $e(0)=\max(|a|,|b|)$, then, following one of the proofs of the Gronwall lemma,
$$
\frac{d}{dx}\ln\left(e(0)+\int_0^x|f(y)|·e(y)\,dy\right)=\frac{|f(x)|·e(x)}{e(0)+\int_0^x|f(y)|·e(y)\,dy}\le |f(x)|
$$
so that after integration
$$
e(x)\le e(0)e^{\int_0^x|f(y)|\,dy}\implies \max(|α(x)|,|β(x)|)\le\max(|a|,|b|)e^{\int_0^\infty|f(y)|\,dy}
$$
