Finding the domain of ODE without calculating a solution We have two ODEs:
  \begin{cases}
    x'(t) = 2x^2-t \\
    x(1) = 1
  \end{cases}
and
  \begin{cases}
    x'(t) = t+e^x \\
    x(1) = 0
  \end{cases}
I am to find(without calculating, only using theorems and slope fields) an interval on which the solution exists.
They both are not Lipschitz so Picard–Lindelöf's doesn't work, but Peano does. We know that a solution exists in $[1 - \eta, 1 + \eta]$. I drew a slope field which seems to be very steep for points outside $[-2, 2]$ for the first one and very steep for $x > 1$ for the second one. 
What can I do next? 
 A: For the 1st question, near $t=1$, $x'(t)\ge 0$. This implies $2x^2(t)\ge t$ or $\frac{t}{x^2}\le \frac12$, $t\in[1,t_0)$ for some $t_0>1$.
So
$$ x'=x^2-t=x^2\bigg[1+(1-\frac{t}{x^2})\bigg]\ge x^2\bigg[1+\frac12\bigg]=\frac{3}{2}x^2. $$
Also
$$ x'=x^2-t\le x^2-1 $$
and hence
$$ \frac32x^2\le x'\le x^2-1. $$
Using $x(1)=1$, it is easy to obtain
$$ \frac{2}{5-3t}\le x(t)\le \frac1{\sqrt2}\frac{(3+\sqrt2)+e^{2\sqrt2(t-1)}}{(3+\sqrt2)-e^{2\sqrt2(t-1)}}. $$
This means that $x(t)$ has a singular point in $(1+\frac{1}{2\sqrt2}(3+\sqrt2), \frac{5}{3})\approx(1.62323, 1.66667)$ and therefore the maximal interval of existence of solution is $[1,t_0)$ with $t_0\in(1.62323, 1.66667)$.
You can use the same way to treat the second equation.
A: Just to check, one comes closer to the solutions by parametrizing in (1) $x=-\frac{u'}{2u}$ to get $u''-2tu=0$, $u(1)=1$, $u'(1)=-2$ which is an Airy equation (linear with smooth coefficients, solution domain $\Bbb R$). As $u''(1)=2$, one can expect (this is no proof) that the function stays convex for a while so that there is no root of $u$ and thus no pole of $x$ in the interval $[\frac12,\frac32]$ around $t=1$.
In (2), set $v=e^{-x}$. Again, a root of $v$ gives a pole of $x$. The equation for $v$ is $v'+tv=-1$, $v(1)=1$ so that $e^{t^2/2}v(t)=1-\int_1^te^{s^2/2}ds\ge 1-(t-1)e^{t^2/2}$ for $t\ge 0$. Numerically one can confirm that there is no root of $v$ in $[\frac23,\frac43]$.
Again, this is not in the spirit of the question, for that one would have to find upper bounds of the solutions, to complement the lower bounds found in the other answer, of the problems as given and the domain where both lower and upper bound exist. This domain is then a subset of the maximal domain of the given problem.
