Show that the following ode has a maximum of one solution.
$$x'(t) = |x(t)| +\sin(x^2(t)+t^2) \quad \text{with} \quad x(0)=0$$
and that this solution satisfy $x(t)\leq e^t-1$ for $t \geq 0$
We got the hint: compare with $x'(t)= |x(t)|+1$
I solved the ODE of the hint and got for $x(t) \geq 0 $ the solution $ x(t)=c_{1}e^t-1 ,c_{1} \in \mathbb{R}$
for $x(t) < 0 $ the solution $ x(t)=c_{2}e^{-t}+1 ,c_{2} \in \mathbb{R}$
Now I´m not sure what to do. I know that if $c_{1}=1$ I get $x(t)\leq e^t-1$ for $t \geq 0$.
What have i to do next? I have problems to show Lipschitz-continuity, because i cant find an L. I need some help.
From $f(t,x)=|x|+\sin(x^2+t^2)$ you get the bound for the difference of the right side under variations of the state variable as \begin{align} |f(t,y)-f(t,x)|&\le\Bigl||y|-|x|\Bigr|+2\sin(\frac{|y^2-x^2|}2)\cos(t^2+\frac{x^2+y^2}2) \\& \le|y-x|+|y-x|\,|y+x| \end{align} so that on the region $|x|\le R$ one gets a Lipschitz constant $L=1+2R$. Until the solution leaves such a region, you get uniqueness. For a global result you need to exclude divergence to infinity in finite time, which the bound on the solution delivers.