Type of differential equation Anyone can give me a hint on how to solve $y(t)'=e^{-t}+y(t)^2$ with $y(0)=0$?
I've got to find out if this takes the Maximal interval of existence in $(T_{-}, T_{+})$ with $T_{+}< \infty$ or $T_{+}= \infty$.
 A: Edit: Let's adress the second part. We have $y(t)'= e^{-t}+ y(t)^2 >0$, hence $y$ is strictly increasing. We set $T=T_+/2$ if $T_+<\infty$ or $T=1$ otherwise. Set $y_1= y(T)>0$ and
$$ x(t) = \frac{1}{\frac{2}{y_1}+T - t} $$ 
for $t\in [0, \frac{2}{y_1}+T)$. On checks that we have
$$ \begin{cases} x(t)' = x(t)^2 \\ x(T) = \frac{y_1}{2} \end{cases}.$$
We define
$$ T_0:= \inf \left\{ t \in \left[1, \frac{2}{y_1}+T \right) \  : \ y(t) > x(t) \right\}.$$
Assume that $T_0 < +\infty$. By continuity of both $x$ and $y$ we have $x(T_0)=y(T_0)$. However, we compute
$$ y(T_0) = y(T) + \int_T^{T_0} y(t)' dt = y(T) + \int_T^{T_0} e^{-t} + y(t)^2 dt  > x(T) + \int_T^{T_0} x(t)^2 dt = x(T) + \int_T^{T_0} x(t)' dt = x(T_0).$$
Hence, $T_0= +\infty$. This means $y(t) > x(t)$ for all $t \in \left[1, \frac{2}{y_1}+T \right)$. But $\lim_{t\uparrow \frac{2}{y_1}+T} x(t) = +\infty$, hence $$\lim_{t\uparrow \frac{2}{y_1}+T} y(t) = +\infty.$$
Thus, $T_+ < +\infty$.
A: To address your very first question:

If you want to solve for an exact and general solution, notice that this is a Riccati Equation.
$$y'=q_0(t)+q_1(t)y+q_2(t)y^2$$
Thus, if we substitute $y=-\frac{u'}{u}$, we can reduce this to a second-order linear ODE:
$$\left(-\frac{u'}{u}\right)'=e^{-t}+\left(-\frac{u'}{u}\right)^2$$
$$-\frac{u''}{u}+\left(-\frac{u'}{u}\right)^2=e^{-t}+\left(-\frac{u'}{u}\right)^2$$
$$u''+e^{-t}u=0$$
Unfortunately, the solution to this ODE contains Bessel Functions of the first and second kind:
$$u(t)=c_1 J_0\left(2\sqrt{e^{-t}}\right)+c_2 Y_0\left(2\sqrt{e^{-t}}\right)$$
This can be differentiated w.r.t. $t$ to obtain an explicit solution for $y(t)$ by substituting into $y(t)=-\frac{u'(t)}{u(t)}$.

Alternatively, if you want to avoid non-elementary functions, you can obtain a series solution to the original ODE.
