Show that a IVP has a unique solution for all $\mathbb{R}$ 
Show that the $$x'(t)=e^{-x^2} + 2x +1 \quad x(0)=x_0 $$ has a unique solution for all $x_0,t \in \mathbb{R}$.

Notes:
Let $f(x)=e^{-x^2} + 2x +1$.
My problem is not showing that there exists a unique solution on a interval, but that there exists a solution for all of $\mathbb{R}$. I know that if I can show that $f(x) $ or $x(t)$ is bounded for all $\mathbb{R}$. But how do I show this when $\lim_{x\to \infty} e^{-x^2} + 2x +1 =\infty$.
Also $f(x)$ is lipschitz continuous with lipschitz constant $2$ as $|e^{-x^2}-e^{-y^2}+2(x-y)|\leq 2|x-y|$.
 A: Probably there are several ways of confirming this fact. Let me suggest a somewhat explicit estimates that actually bound the solution $x(t)$. 
Observe that $0 \leq \, e^{-x^2} \leq 1$. Take a local solution $x = x(t)$ of the IVP
$$\frac{dx}{dt} = e^{-x^2} + 2 \, x + 1$$
$$x(0) = x_0$$ for any $x_0 \, \in \mathbb{R}$. Then 
$$1+ 2 \, x(t) \, \leq \,\frac{dx}{dt}(t) = x'(t) = e^{-x(t)^2} + 2 \, x(t) + 1 \,  \leq \, 2 + 2\, x(t)$$ Define the function $$z(t) = e^{-2t} x(t) + e^{-2t}$$ Then $$z'(t) = \frac{d}{dt}\left(e^{-2t} x(t) + e^{-2t}\right) = e^{-2t} x'(t) - 2 e^{-2t} x(t) - 2 e^{-2t} = e^{-2t}\big(x'(t) - 2 \, x(t) -2 \big)$$
$$= e^{-2t}\Big(e^{-x(t)^2} + 2\, x(t) + 1 - 2 \, x(t) -2 \Big) = e^{-2t}\Big(e^{-x(t)^2} - 1 \Big) \leq 0 $$ Therefore, $z(t)$ is non-increasing function and hence for $0 \leq t$ in the domain of $x(t)$ we have that $$z(t) =  e^{-2t} x(t) + e^{-2t} \leq z(0) = x_0 + 1$$ i.e.
$$x(t) \leq (x_0 +1) e^{2t} - 1$$ Analogously, define $$y(t) =  e^{-2t} x(t) + \frac{1}{2} e^{-2t}$$
Then $$y'(t) = \frac{d}{dt}\left(e^{-2t} x(t) + \frac{1}{2}e^{-2t}\right) = e^{-2t} x'(t) - 2 e^{-2t} x(t) - e^{-2t} = e^{-2t}\big(x'(t) - 2 \, x(t) -1 \big)$$
$$= e^{-2t}\Big(e^{-x(t)^2} + 2\, x(t) + 1 - 2 \, x(t) - 1 \Big) = e^{-2t}\Big(e^{-x(t)^2}\Big) \geq 0 $$ Therefore, $y(t)$ is non-decreasing function and hence for $0 \leq t$ in the domain of $x(t)$ we have that $$y(t) =  e^{-2t} x(t) + \frac{1}{2}e^{-2t} \geq y(0) = x_0 + \frac{1}{2}$$ i.e.
$$x(t) \geq \left(x_0 +\frac{1}{2}\right) e^{2t} - \frac{1}{2}$$ Consequently, for $t \geq 0$ within the interval of definition of $x(t)$ the following inequalities hold
$$\left(x_0 +\frac{1}{2}\right) e^{2t} - \frac{1}{2} \,\leq \, x(t) \, \leq \, (x_0 +1) e^{2t} - 1$$ Since the explicit functions on both sides of the inequality are defined for all $t \geq 0$ the function $x(t)$ should also be defined for all $t \geq 0$. 
Absolutely analogous argument shows that $x(t)$ should exist for $t \leq 0$. 
Edit. Observe that one can do absolutely the same thing (literally the same argument) if one is given any function $\phi(t,x) \, \geq 0$. Then one can look at the quite complicated and most likely explicitly unsolvable  differential equation 
$$\frac{dx}{dt} = e^{- \phi(t,x)} + 2 \, x + 1$$ and observe again that $$0 \leq \, e^{-\phi(t,x)} \, \leq 1$$ The only think that matters here is the linear part of the equation $2 \, x + 1$ and the consequent linear bounds 
$$2 \, x + 1 \, \leq \,\frac{dx}{dt} = e^{- \phi(t,x)} + 2 \, x + 1 \, \leq \, 2 \, x + 2$$ Everything goes from here unaltered.
As a generalization, one can prove the same result, using almost the same technique, for the general case 
$$\frac{dx}{dt} = f(t, x)$$ for 
$$a \, x + b \leq \,f(t,x) \leq A \, x + B$$
In this case the functions are for instance $$z(t) = e^{-At} \, x(t) + \frac{B}{A} \, e^{-At}$$ and $$y(t) = e^{-at} \, x(t) + \frac{b}{a} \, e^{-at}$$
