Differential equation: $\ddot{y}(x) + \alpha\dot{y}^2(x) + \beta y(x) = 0$ I am interested in finding an approximate solution for this differential equation, since the exact analytic solution seems to not exist. I tried with Mathematica and it spits out nothing.
$$\ddot{y}(x) + \alpha\dot{y}^2(x) + \beta y(x) = 0 ~~~~~~~ (\alpha, \beta) > 0$$
Clearly, I don't want an approximate solution like "let's cut one of the three terms". I was thinking about a good ansatz, but I found nothing really good. Also I got stuck while trying
$$y(x) = A(x)\ e^{-kt}$$
Sounds a really "simple" problem but maybe it's not. Any idea?
 A: I want to expand upon Brevan Ellefsen's comment and my comments a little.
I assume that dots in OP's case mean $x$ derivative, since $y(x)$ depends only on $x$. Then we can easily get rid of $\alpha$ and $\beta$.
$$
\frac{d^2 y}{dx^2}+\alpha \left(\frac{d y}{dx} \right)^2+\beta y=0
$$
Replacing:
$$
x \rightarrow \frac{x}{\sqrt{ \beta }}~~~~~y \rightarrow \alpha y
$$
We will get.
$$
\frac{d^2 y}{dx^2}+\left(\frac{d y}{dx} \right)^2+y=0
$$
Here it's recommended to make a replacement:
$$
w(y)=\left(\frac{d y}{dx} \right)^2
$$
Then:
$$
\frac{d^2 y}{dx^2}=\frac{d \sqrt{w}}{dx}=\frac{d \sqrt{w}}{dy} \sqrt{w}=\frac{d w}{2dy}
$$
The equation transforms to:
$$
\frac{d w}{dy}+2 w+2 y=0
$$
We only need to solve two first order ODEs. The solution for $w$ can be found easily (I used Wolframalpha):
$$
w(y)=C_1 e^{-2y}-y+\frac{1}{2}
$$
Then we need to solve first order nonlinear equation:
$$
\left(\frac{d y}{dx} \right)^2=C_1 e^{-2y}-y+\frac{1}{2}
$$
$$
\frac{d y}{dx}=\pm \sqrt{C_1 e^{-2y}-y+\frac{1}{2}}
$$
I suggest a simple numerical scheme. Also, since this equation is fully separable, we can try to approximate the integral instead. 
For $C_1=0$ this ODE is exactly solvable.
$$
y=\frac{1}{2}-\left(\frac{x}{2}+C_2 \right)^2
$$
This is the exact partial solution of the original equation, which can be checked by direct substitution.
Or, alternatively if $|C_1|>>1$ and also much larger than maximum value for $y$, we can set $c_1=\sqrt{C_1}$ and get:
$$
\frac{d y}{dx} \approx \pm c_1 e^{-y}
$$
$$
y \approx \ln (|c_3 x|)
$$
This is a solution of the original equation without the third term (or with $\beta \rightarrow 0$).
And this is the answer to OP's confusion - since $y(x)$ can be approximated by $\ln(x)$ it's hardly surprising that their exponential ansatz did not fit well.

Some further ramblings:

 As for the best way to approximate the solution - it heavily depends on the context - if it was a simple pendulum in air for example, then you have harmonic oscillator motion with resistance proportional to velocity squared, and the second term can be interpreted as perturbation. However, since $\alpha$ is positive, it's apparently some other case. There is also the case of nonlinear Schrodinger, and there is a number of methods developed to solve this problem also.

My point is - there can be no 'general' approximate solutions, the appropriate approximation depends on the context of the original problem.
