Suppose we want to solve a simple ODE $$\dot{x}=2t(1-x)=f(t,x)\qquad (1)$$ with initial condition $x_0=2$ using the contraction mapping principle.
I showed that $f$ is uniformly Lipschitz continuous in $x$ over $[0,1]\times\mathbb R$ for all $t\in[0,1]$ with Lipschitz constant $L=2$.
Considering a normed vector space $(C[0,\delta],||\cdot||_C)$, the contraction map with $P:C[0,\delta]\to C[0,\delta]$ given by $$(P\phi)(t)=x_0+\int_{0}^{t}f(t,\phi(\tau))\;d\tau.$$
Now, by direct integration the solution of $(1)$ is $$x=1+\exp(-t^2).$$
So far we have shown that there exists a unique solution over the interval $[0,\delta]$, right?
Is the next step to provide an argument using the contraction mapping that shows that the solution is unique over the entire set of positive real numbers?
I am not sure how to make this argument. I thought that maybe by iteration and increasing $\delta$ one can show that the solution is unique over a bigger interval. But this doesn't see formal/rigorous.
I'd appreciate any help/hint. Thank you.