Show that the solution to an IVP blows up in finite time I want to show that the solution to the IVP $$x'(t) = t + e^{-t}e^x, \hspace{0.5mm} x(0) = 0 $$ blows up in finite time. That is, I want to find a $T > 0$ such that $x(t)$ is defined on $[0,T)$ and $x(t) \rightarrow \infty$ as $t \rightarrow T^{-}$.
I haven't even managed to find a solution to the IVP. I thought that if I could find the solution, then I could "simply" find a $T$ such that that solution blows up when $t \rightarrow T$.  
The closest I've come to finding a solution is the following: I found that if $x(t) = \frac{t^2}{2} - e^xe^{-t} + h(x)$, then $x'(t) = t + e^{-t}e^x$. And then setting $x(0) = 0$ gives $h(x) = e^x$. So $x(t) = \frac{t^2}{2} - e^{-t}e^x + e^x$ should (?) solve the IVP. If this was indeed the way to go, then how would I go from here to find a $T$ such that $x(t) \rightarrow \infty$?
 A: With that nasty $e^x$ nonlinear term, it suggests setting $x = a\ln u$, then adjusting $a$ to make a linear equation. This works for $a = -1$, so we set $x = -\ln u$ and get
$$
u'(t) + t u(t) = - e^{-t}\;\;\;\;;\;\;\;\; u(0)  = 1.
$$
This first-order equation can be solved with the integrating factor $e^{t^2/2}$, giving
$$
\int_0^t\frac{d}{dt'}\left[e^{t'^2/2}u(t')\right]dt' = e^{t^2/2}u(t) - u(0) = -\int_0^te^{t'^2/2-t'}dt'.
$$
So 
$$
u(t) = e^{-t^2/2}\left[1 - \int_0^te^{t'^2/2-t'}dt'\right].
$$
Since $x = -\ln u$, it blows up when $u(T) = 0$, which happens when
$$
\int_0^Te^{t'^2/2-t'}dt' = 1.
$$
Since $e^{t'^2-t'}$ is an unbounded function, this must happen at some finite $T$ (and indeed, it happens near $T = 1.44$).
A: Set $u=e^{x-t}$. Then $u'=u(t-1+u)$ with $u(0)=1$ is a Bernoulli equation. With the usual substitution $u=v^{-1}$ this transforms into the linear ODE
$$
-v'=(t-1)v+1, ~~ v(0)=1\implies v(t)=e^{-\frac12(t-1)^2}\left[\sqrt{e}-\int_{-1}^{t-1}e^{+\frac12s^2}ds\right]
$$
As the integral term grows to infinity, $v$ will have a root that is then a pole for $u$ and a blow-up point for the solution of the IVP. The root is close to $T\approx 1.4392$
