I am struggling with this problem: for which $y_0\in \mathbb{R}$ does $$y'(t)=(t+y(t))^2,y(0)=y_0$$ have local solutions? For which $y_0$ are these solutions unique?
By Peano, as $f(t,y) = (t+y(t))^2$ is continuous, so there must be local solutions. But how do I find out whether they are unique? I don't even know whether they are.
I tried to check whether $(t+y(t))^2$ is locally Lipschitz, but $$\lVert f(t,y)-f(t,\bar{y})\rVert = \lVert2t(y-\bar{y})+y^2-\bar{y}^2\rVert$$ and I don't see how I can use this expression to check local Lipschitz, as it depends on $t$.
For $t\in[0,A]$ and $\|y\|\le R,\|\bar{y}\|\le R$, one has $$\lVert f(t,y)-f(t,\bar{y})\rVert = \lVert2t(y-\bar{y})+y^2-\bar{y}^2\rVert=\lVert y-\bar y\lVert\lVert2t+y+\bar y\lVert\le L\lVert y-\bar y\lVert$$ with $L=2A+2R$; i.e., the $f(t,y)$ satisfies the Lipschitz condition in $[0,A]\times[-R,R]$. Therefore there is a unique solution.