$f '' - (f ')^2 + f=0$; what is known about solutions? I'm curious about solutions to the equation 
$$f''-(f')^2+f=0$$
on the whole real line, as well as solutions which are periodic. Any info about the obvious multivariable generalization would interest me as well.
I'm not necessarily looking for explicit solutions, though if there are nontrivial explicit solutions that would be interesting. I'm curious about techniques used to analyze such solutions and the general features of the solutions. I'm not well versed in ODE's, but this equation came up in something I was looking at, so don't be afraid to talk down to me and start with the basics if need be.
 A: The most insightful way to study this equation is to write it as a two-dimensional dynamical system, like
\begin{align}
 f' &= g, \\
 g' &= g^2 -f. \tag{1}
\end{align}
Questions about the shape of solutions can best be answered by looking at the orbits of solutions in the $(f,g)$ phase plane. If you consider such an orbit as a graph $g(f)$, then these obey the differential equation
\begin{equation}
 \frac{\text{d} g}{\text{d} f} = \frac{g'}{f'} = g - \frac{f}{g}. \tag{2}
\end{equation}
The solutions of $(2)$ are given by
\begin{equation}
 g(f) = \pm \sqrt{\frac{1}{2}+f + c_0 e^{2 f}}. \tag{3}
\end{equation}
As you can see, these functions $g(f)$ exist as long as $\frac{1}{2}+f + c_0 e^{2 f} \geq 0$. That is, the boundaries of the domain of $g(f)$ are given by the solutions to the equation $\frac{1}{2}+f + c_0 e^{2 f} = 0$, which yields
\begin{equation}
 f = -\frac{1}{2}\left(1+W(2 c_0/e)\right), \tag{4}
\end{equation}
where $W$ is the Lambert W function. This function has two branches, which coincide at the left point of its domain of definition, which is at $-\frac{1}{e}$. This coincides with the observation that $\frac{1}{2}+f + c_0 e^{2 f} < 0$ for all $f$ if $c_0 < -\frac{1}{2}$. Also, the second branch of the Lambert $W$ function only exists as long as its argument lies in between $-\frac{1}{e}$ and $0$. Therefore, we can identify two limit values of $c_0$ (being $-\frac{1}{2}$ and $0$), separating two types of solutions.
For $-\frac{1}{2} < c_0 < 0$, the two graphs $(3)$ form a closed orbit. As $c_0 \to -\frac{1}{2}$, these orbits shrink to the point $(0,0)$ in the $(f,g)$ phase plane, which is an equilibrium of the system $(1)$. As $c_0 \to 0$, these periodic orbits grow larger and larger (in the phase plane), until they become unbounded. For $c_0 = 0$, the graph is given by the curve
\begin{equation}
 g(f) = \pm \sqrt{\frac{1}{2} + f}. \tag{5}
\end{equation}
Remembering that $g = f'$, this yields a differential equation
\begin{equation}
 f'^2 = 1 + f,
\end{equation}
which for $f(0) = 0$ yields
\begin{equation}
 f(t) = \frac{t^2}{4} \pm \frac{t}{\sqrt{2}}. \tag{6}
\end{equation}
For all values $c_0>0$, the graph $(3)$ is unbounded, and hence the orbit on that graph is unbounded as well. In the phase plane, the curves $(5)$, which are traced out by the orbit of $(6)$, form a separatrix between the region in phase space where closed orbits exist ('within' the parabola), and where orbits are unbounded.
To calculate the period $T(c_0)$ of a periodic orbit for a certain $-\frac{1}{2} < c_0 < 0$, the graph of which exists between the two solutions $(4)$ (let's denote these by $f_\text{left}$ and $f_\text{right}$), you have to evaluate the integral
\begin{equation}
 T(c_0) = 2\int_{f_\text{left}}^{f_\text{right}} \frac{1}{\sqrt{\frac{1}{2}+f + c_0 e^{2f}}}\,\text{d} f,
\end{equation}
which does not have a closed form expression. 
For $c_0 > 0$, the large-$f$ behaviour of $(3)$ can be approximated as $g(f) \leadsto \sqrt{c_0} e^f$ as $f \to \infty$. This implies that for large $f$, we can write
\begin{equation}
 f(t) \leadsto - \log \sqrt{c_0} |t| \quad \text{as}\quad t \to \pm \infty.
\end{equation}
I strongly advise you to draw the phase plane associated to $(1)$, to get more insight into the behaviour of the system. You can use PPlane for that.
A: The ODE is on the "autonomous" kind :http://mathworld.wolfram.com/Autonomous.html
The usual method to reduce the order is applied below.
The result is the inverse function, on integral form. There is no closed form.
Note : For easiness the symbol $f$ is replaced by $y$.

