Using bounds for an analytically intractable differential equations problem Exercise 2.8.6 from Strogatz's Nonlinear Dynamics and Chaos with Applications asks us to consider the initial value problem $\dot x = x+e^{-x}$, $x(0)=0$ $\left(\dot x\, \text{means}\, \frac{dx}{dt}\right)$, which cannot be solved analytically. 
Part (b) of the exercise states as follows: 

Using some analytical arguments, obtain rigorous bounds for the value of $x$ at $t=1$. In other words, prove that $a<x(1)<b$ for $a$, $b$ to be determined. By being clever, try to make $a$ and $b$ as close together as possible.

I was given a hint from my professor that, in other words, we want to approximate $$\psi(x) \leq x+ e^{-x} \leq \phi(x) $$ for $\phi(x),\,\psi(x) \geq 0$, and that essentially what we want to do is solve the following three IVPs:
$$(1) \qquad\frac{dx}{dt} = \psi(x) \\ \qquad x(t_{0})=x_{0}, $$
$$ (2) \qquad  \frac{dx}{dt} = x+e^{-x} \\ \qquad x(t_{0}) = x_{0},$$
and 
$$ (3) \qquad \frac{dx}{dt} = \phi(x) \\ \qquad x(t_{0})=x_{0} .$$
Then, if we call the solution to $(1)$ $a(t)$, the solution to $(2)$ $x(t)$, and the solution to $(3)$ $b(t)$, the problem boils down to showing that $$a(1) \leq x(1) \leq b(1). $$
In that regard, I am very close to finding $b(1)$. To get there, I considered the Taylor expansion for $$e^{-x} = \sum_{n=0}^{\infty} (-1)^{n} \frac{x^{n}}{n!} =1-x + \frac{x^{2}}{2} - \frac{x^{3}}{3!}+ \cdots$$ so that $x+e^{-x} = x + \sum_{n=0}^{\infty} (-1)^{n}\frac{x^{n}}{n!}.$ As a bound, I decided to show that $x+e^{-x} \leq 1 + \frac{x^{2}}{2}$ $\forall x \geq 0.$
To that effect, I let $h(x) = x + e^{-x} - 1 - \frac{x^{2}}{2}$, and then I considered $h^{\prime}(x) = 1 - e^{-x} - x$ and $h^{\prime \prime}(x) = e^{-x} - 1$.
Since $h^{\prime \prime}(x) > 0$ for $x < 0$, $h^{\prime}(x)$ is increasing on $(-\infty, 0)$. 
Since $h^{\prime \prime}(x) < 0$ for $x > 0$, $h^{\prime}(x)$ is decreasing on $(0, \infty)$. 
But, $h^{\prime}(x) = 0$ for $x = 0$, so $h^{\prime}(x) < 0$ for $x<0$ and $h^{\prime}(x) < 0 $ for $x > 0$ as well - the only time $h^{\prime}(x)$ is not negative is when $x=0$, and there it is simply $=0$.
Therefore, $h$ is everywhere decreasing, and since $h(x)=0$, $h > 0$ for $x<0$ and $h < 0$ for $x > 0$, so we have that $h(x) = x + e^{-x} - 1 - \frac{x^{2}}{2} \leq 0$ for $x \geq 0$, which implies that $x + e^{-x} \leq 1 + \frac{x^{2}}{2} $, as desired. 
So, then I let $\phi(x) = 1 - \frac{x^{2}}{2}$, and I need to solve Equation $(3)$ above. 
Using separation of variables, I have that $\frac{dx}{dt} = 1 - \frac{x^{2}}{2}$ becomes $\displaystyle \int \frac{1}{\displaystyle 1 - \left(\frac{x}{\sqrt{2}} \right)^{2}} dx = \int dt$, which comes out to be 
$$\ln \left(1 + \frac{x}{\sqrt{2}} \right) - \ln \left(1 - \frac{x}{\sqrt{2}} \right) = \sqrt{2}t + C, $$
And then, exponentiating both sides, we get $$\frac{1+\frac{x}{\sqrt{2}}}{1-\frac{x}{\sqrt{2}}}= Ce^{\sqrt{2}t} $$
From here, I don't know what to do. The problem tells us what the initial condition when $t=0$ is - it tells us that $x(0) = 0$, but it doesn't tell us what it is for $t=1$. And since we're trying to find bounds for $x(1)$, I'm assuming that for Equation (3) (actually in all three of the equations), we want $t_{0} = 1$, right? The only problem is that I can't figure out what $x_{0}$ is in that case, so how do I solve for $C$?
Also, if you could help me out on the left bound (i.e., solving for $\psi(x)$), I would appreciate that very much!
Thank you ahead of time for your help.
 A: As @Evgeny proposed use the inequality
$$x \leq \dot{x} \leq x + 1.$$
Then apply the following Theorem (Petrovitsch, 1901): 

If $u$ satisfies the differential inequality $u′(t)≥f(u(t),t)$, and $y$ is
  the solution to the ODE $y′(t)=f(y(t),t)$ under the boundary condition
  $u(t_0)=y(t_0)$, then: $\forall t<t_0, u(t)\leq y(t) \,\, \forall t>t_0, u(t)\geq y(t)$.

A: As $e^x\ge 1+x$, you get $\dot x\ge 1$, which is easier to solve than your lower bound.
In your lower bound computation, inserting $x(0)=0$ gives $C=0$. Then solving for $x$ is just arithmetic/algebraic manipulations. 
For the upper bound, as long as $x<3$ we get by the Leibniz test $e^{-x}\le 1-x+\frac12x^2$. You showed that this is even valid for the whole positive half axis. Thus
$$
\dot x\le 1+\frac12x^2
$$
Both differential inequalities can be solved via standard integration methods an give an interval of length about $0.2$ for $x(1)$.
One can improve the lower bound to get a similar formula as for the upper bound by using $$e^{-x}=(e^{-x/2})^2\ge\Bigl(1-\frac x2\Bigr)^2$$ to get
$$
\dot x\ge 1+\frac14x^2
$$
As we know that $\dot y=1+Ay^2$ has the solution $y(t)=\frac1{\sqrt A}\tan{\sqrt{A}t}$, you can get the bounds
$$
2\tan\frac12\le x(1)\le\sqrt2\tan\frac1{\sqrt2}
$$
which falls between $1.09$ and $1.21$.
