An application of Banach fixed point theorem for initial value problem Find a condition for $\beta>0$ which implies that the differential equation system  \begin{align}
x'(t)&=x(t)+y(t) ,\\ y'(t)&=t^{2}+tx(t)
\end{align}
 with initial conditions $x(0)=0, y(0)=0$ has a unique solution $(x(t),y(t))$ in $C[-\beta,\beta]\times C[-\beta,\beta]:=X$
My attempt:
The metric on $X$ is defined by $d((a,b),(c,d))=d_{1}(a,c)+d_{1}(b,d)$ where $d_{1}$ is the sup-metric on $C[-\beta,\beta] $ and I use the Banach fixed point theorem for the map $T:X \to X$ by 
$$
T(x(t),y(t))=\left(\int_{-\beta}^{\beta}(x(t)-y(t))dt,\,\int_{-\beta}^{\beta}(t^{2}+tx(t))dt \right)$$
Say $u=(x(t),y(t))$ and $v=(x_{1}(t),y_{1}(t))$
To see that $T$ is a contraction, I consider 
$$
d(Tu,Tv)=sup|\int_{-\beta}^{\beta}[x(t)-y(t)-x_{1}(t)+y_{1}(t)]dt|+sup|\int_{-\beta}^{\beta}t(x(t)-x_{1}(t))dt|
$$ 
How can we complete the proof? Thanks!
 A: You should consult your standard proof of the Picard-Lindelöf theorem to find the formula for interval length. 
In general terms, the size of the interval is external to the ODE, as a linear ODE with continuous coefficient matrix the solution can be extended to all of $\Bbb R$. Which in turn means that the interval you get with the standard proof depends on details of the computation, you have to present them very carefully. 
The interval independence of the contractivity can be shown by an adapted variant of the usual proof. As usual, transform the IVP
$$
y'(t)=A(t)y(t)+b(t),\quad y(0)=y_0
$$
into the integral form
$$
y(t)=T(y)(t)=y_0+\int_0^t \bigl(A(s)y(s)+b(s)\bigr)\,ds
$$
Fix any interval $[−β,β]$, set
$$
L=\sup_{t\in[−β,β]}\|A(t)\|
$$
for your favorite choice of vector norm and induced operator norm, and show contractivity of the integral operator with respect to the modified supremum norm
$$
\|y\|_L=\sup_{t\in[−β,β]}e^{-2L|t|}\|y(t)\|
$$
Because then you get for two different continuous functions $y$ and $\bar y$ with $y(0)=y_0=\bar y(0)$ (and assuming for the moment $t>0$, the other case is analogous)
\begin{align}
e^{-2Lt}\|T(y)(t)-T(\bar y)(t)\|
&\le
\|y(0)-\bar y(0)\|+e^{-2Lt}\,\int_0^t\|A(s)\|\,\|y(s)-\bar y(s)\|\,ds
\\
&\le
e^{-2Lt}\int_0^tLe^{2Ls}\|y-\bar y\|_L\,ds=\frac{1-e^{-2Lt}}{2L}\,L\,\|y-\bar y\|_L
\\
\implies
\|T(y)-T(\bar y)\|_L&\le
\frac12\|y(s)-\bar y(s)\|_L
\end{align}
As you see, by an appropriate choice of the function space norm one can get contractivity with factor $1/2$ for any given interval.
