u'=u^2
u(0)=1
find $\displaystyle \delta$>0 s.t. the solution exists $\displaystyle [-\delta,\delta]$
hint?
The DE is of the type 'with separable variables' , so that its solution can be found in 'standard way' ...
$\displaystyle \frac{du}{dt} = u^{2} \rightarrow \frac{du}{u^{2}} = dt \rightarrow \frac{2}{u} = c - t$ (1)
... and taking into account the 'initial conditions' we have...
$\displaystyle u= \frac{1}{1-\frac{t}{2}}$ (2)
It is evident from (2) that u(*) has a singularity in $\displaystyle t=2$ and this fact is confirmed if we expand u(*) in series around $\displaystyle t=0$ ...
$\displaystyle u(t) = \sum_{n=0}^{\infty} (\frac{t}{2})^{n}$ (3)
... which converges for $\displaystyle |t|<2$. So the value of $\displaystyle \delta$ You are searching is $\displaystyle \delta=2$...
Kind regards
$\displaystyle \chi$ $\displaystyle \sigma$