Asymptotic behaviour / non-linear ODE I'm trying to find the behaviour at large $t$ of solutions to the non-linear differential equation:
$$\dfrac{d^2y}{dt^2}-\left(\dfrac{dy}{dt}\right)^2+(2-a)t\dfrac{dy}{dt}+2ay=0$$
I tried to replicate the approach detailed on Wikipedia, making the assumption that $y(t)\sim e^{S(t)}$ as $t\to\infty$ for some function $S(t)$, but the $\left(\dfrac{dy}{dt}\right)^2$ term in the DE leaves me with an extra factor of $e^{S(t)}$ that I don't know how to deal with.
How do I go about finding the asymptotic series for $y(t)$?
 A: The assumption $y \sim \exp S$ is the way to go if the highest derivative is multiplied with a small factor.
For your problem, the correct approach is to single out the (two or more) terms that dominate the others for $t\to\infty$. Which terms this are is a bit trial and error (and some experience).
You did not specify what values of $a$ you are interested in (the analysis depends on that). I assume for the following that $0<a<2$. If you are interested in other values, please just comment and I will edit the answer.
We claim that the solution to $$ (2-a) t y'(t) +2a y(t)=0$$
gives the correct asymptotic expression for $t\to\infty$. We obtain the solution
$$y(t) \sim c t^{-a/(1-a/2)}.$$
We observe that 
$$ \lim_{t\to\infty} \frac{y'^2}{y} = \text{const} \lim_{t\to\infty} t^{-4/(2-a)}=0  $$
and 
$$ \lim_{t\to\infty} \frac{y''}{y} = \text{const} \lim_{t\to\infty} t^{-2}=0  $$
and thus the remaining terms are subdominant and we have obtained the correct asymptotic solution.
Appendum:
Due to the request of the OP, I add the analysis for $a>2$ in the following: this is a untypical case as we need to keep three terms of the differential equation. We note that with $y(t) \sim \alpha t^2$, we have that $y''(t) \ll y(t)$. Thus the differential equation is given by
$$0 =-y'^2 +(2-a) t y'  +2a y = 4 (1-\alpha) \alpha t^2.$$
With the solutions $\alpha_0\in\{0,1\}$ (only $\alpha_0=1$ makes sense as otherwise the solution is not behaving like $t^2$)
We try to obtain a better estimate and substitute $y(t) = t^2  + \epsilon(t)$. We obtain the new equation
$$2 + 2a \epsilon -(2+a) t \epsilon'- \epsilon'^2 + \epsilon''  =0.$$
The dominating terms for $t\to\infty$ are given by
$$2a \epsilon-(2+a) t \epsilon' =0$$
which can be explicitly checked. The solution (for $a>2$) is thus asymptotically given by
$$ y(t) \sim t^2 + c t^{a/(a/2+1)}$$
with an constant $c$.
The asymptotic expansion is of course only valid for a certain set of initial conditions such that the solution does not diverge before. From numerics (see Robert Israels anwer), it seems that the solution is unstable; meaning that closeby trajectories will actually move away from the asymptotic result given above. As such, the curve $y(t)=t^2$  seems to serve as a separatrix.
However to see that the result is correct, it is best to integrate the equation backwards  (from large to small times). Below, I give results for the initial problem
$$ y(10^3) =10^6 +10^5, \qquad y'(10^3) = 0$$
close to $y(t)\sim t^2$.
We have (blue numerical result, orange $t^2$)

Subtracting $t^2$, we obtain the subdominant term $t^{a/(a/2+1)}=t^{6/5}$
(blue $y_\text{numeric}(t) -t^2$, orange $10 t^{6/5}$)

A: Not a whole answer: The combination $y'(-y'+(2-a)t)$ suggests that a quadratic function might play some role; in fact $y(t) = t^2-a^{-1}$ is one solution when $a\ne0$. Maybe you can vary from that.
A: For $a=3$, numerical solutions seem to indicate solutions that blow up at a finite time.  Here's a typical plot.

In such a case, there is no $t \to \infty$ to study.
Note that the equation $y'' - (y')^2 = 0$, which can be solved in closed form,
does have solutions of this type: $y(t) = -\ln(c (t - t_0))$.
A: Case $1$: $a=2$
The ODE becomes $\dfrac{d^2y}{dt^2}-\left(\dfrac{dy}{dt}\right)^2+4y=0$ , which belongs to a second-order autonomous ODE
Let $u=\dfrac{dy}{dt}$ ,
Then $\dfrac{d^2y}{dt^2}=\dfrac{du}{dt}=\dfrac{du}{dy}\dfrac{dy}{dt}=u\dfrac{du}{dy}$
$\therefore u\dfrac{du}{dy}-u^2+4y=0$
$\dfrac{1}{2}\dfrac{d(u^2)}{dy}-u^2=-4y$
$\dfrac{d(u^2)}{dy}-2u^2=-8y$
$\dfrac{d(e^{-2y}u^2)}{dy}=-8ye^{-2y}$
$e^{-2y}u^2=(4y+2)e^{-2y}+C_1$
$\left(\dfrac{dy}{dt}\right)^2=C_1e^{2y}+4y+2$
$\dfrac{dy}{dt}=\pm\sqrt{C_1e^{2y}+4y+2}$
$dt=\pm\dfrac{dy}{\sqrt{C_1e^{2y}+4y+2}}$
$t=\pm\int^y\dfrac{dy}{\sqrt{C_1e^{2y}+4y+2}}+C_2$
Case $2$: $a=0$
The ODE becomes $\dfrac{d^2y}{dt^2}-\left(\dfrac{dy}{dt}\right)^2+2t\dfrac{dy}{dt}=0$ , which belongs to the ODE of the form http://eqworld.ipmnet.ru/en/solutions/ode/ode0347.pdf
$\dfrac{d^2y}{dt^2}=\left(\dfrac{dy}{dt}\right)^2-2t\dfrac{dy}{dt}$
$\dfrac{\dfrac{d^2y}{dt^2}}{\dfrac{dy}{dt}}=\dfrac{dy}{dt}-2t$
$\ln\dfrac{dy}{dt}=y-t^2+c$
$\dfrac{dy}{dt}=c_1e^ye^{-t^2}$
$e^{-y}~dy=c_1e^{-t^2}~dt$
$\int e^{-y}~dy=c_1\int e^{-t^2}~dt$
$-e^{-y}=c_1\int_0^te^{-t^2}~dt+c_2$
$e^{-y}=C_1\int_0^te^{-t^2}~dt+C_2$
