Where did your last expression come from? When searching for a series solution you are supposed to explicitly write power series on both sides and carefully match coefficients.
Let's assume, for small $t$, the function can be written as:
$$y(t)=a_0+a_1t+a_2 t^2+ \cdots$$
Now we substitute this into the equation.
Up to first order we have: $$2a_2+6 a_3 t+a_0 t=1$$ Up to second order: $$2a_2+6 a_3 t+12 a_4 t^2+a_0 t+a_1 t^2=1-\frac{1}{2} t^2$$ and so on. Now we match the coefficients for the same powers of $t$. We can see here that: $$2a_2=1 \\ 6a_3+a_0=0 \\ 12 a_4+a_1=-\frac{1}{2}$$
If there's a general pattern, we can try to derive it. It's not always clear from the start, which is why I prefer writing down the first few terms explicitly.
In any case, I don't see what kind of trouble you had with your general expression? Yes, series for $\cos$ have only even powers. That just makes the equations easier to solve.
Note that we can write the cosine series as:
$$\cos t=\sum_{n=0}^\infty \frac{(-1)^n t^{2n}}{(2n)!}=\sum_{n=0}^\infty \frac{c_n t^n}{n!}$$
Where $c_{2k}=(-1)^k$ and $c_{2k+1}=0$, for $n=2k$ or $n=2k+1$.
Can you match the coefficients for your equation now?
Update
Another way would be to use the standard methods of solving inhomogeneous linear ODEs.
We search for a particular solution $y_p$, and write the general solution as a sum of the general solution of homogeneous equation $y_h$ and $y_p$.
$$y(t)=y_h(t)+y_p(t)$$
$$y_h(t)=C_1 \text{Ai} (-t)+C_2 \text{Bi} (-t)$$
Searching for $y_p$ might be done in various ways, for example, search in the form:
$$y_p(t)=C(t) \text{Ai} (-t)$$
Then:
$$C''\text{Ai} (-t) -2 C' \text{Ai}' (-t)= \cos t$$
Set:
$$C'(t)=f(t)$$
$$\text{Ai} (-t) f'-2 \text{Ai}' (-t) f= \cos t$$
This is a first order linear equation and can be solved in the usual way.
Of course, this method leads to some very awkward integrals, but it gives exact general solution in a manner of speaking.
With regard to the first question:
It's not so much the rapid oscillations themselves, but rather that the area associated with each oscillation decreases as $\lvert x\rvert \rightarrow \infty $.
Considering the real part of the integral:
$$ \Re(I) = \int_{-\infty}^{\infty} \cos{(x^3/3+sx)}dx = 2\int_{0}^{\infty} \cos{(x^3/3+sx)}dx$$
This integral can be realized as a summation of a sequence $(a_n)$, where $a_n$ is the area of each half-oscillation:
$$ \Re(I)=2\sum_{n=0}^{\infty} a_n$$
For this summation:
- $a_n$ will alternate between positive and negative values
- You can show that $a_n \rightarrow 0$
- $\lvert a_n \rvert$ is decreasing
And hence by the alternating series test, the summation converges
Best Answer
The function you are given comes from distorting a contour of integration using Calculus of residues technics. In the reference I gave you in the comments, they use a contour that starts at the origin to a point in the positive $x$-axis at a distance $R$, then a moving along the circle of radius $R$ up to a point at an angle $\theta$, and then moving back to the origin along the line parallel to $e^{i\theta}$. After that, they let $R\rightarrow\infty$.
There are many other ways to distort the contour for the Airy function. In any case, for the problem in the OP you don't need to identify which distortion is used. It suffices to differentiate and see that the function given to you satisfies Airy's equation.
Notice that the function, as is given to you, is actually analytic in $x$ since the term $e^{-t^3/3}$ decays to $0$ faster than any polynomial. That means that one can exchange the order of differentiation and integration without any qualms (the justification being dominated convergence).
Let $ F(x)=\int_{0}^{\infty}\omega\exp(-\frac{t^3}{3}+ix\omega t)dt$ , where $\omega = e^{\frac{i\pi}{6}}$
$$F'(x)=\int^\infty_0\exp(-\tfrac{t^3}{3})i\omega^2 t\exp(ix\omega t)\,dt$$
$$\begin{align} F''(x)&= -\int^\infty_0\exp(-\tfrac{t^3}{3})\omega^3 t^2\exp(ix\omega t)\,dt\\ &=-i\int^\infty_0\exp(-\tfrac{t^3}{3}) t^2\exp(ix\omega t)\,dt \end{align} $$
Integration by parts yields $$\begin{align} x\,F(x)&=-i\int^\infty_0 \exp(-\tfrac{t^3}{3})i\omega x\exp(ix\omega t)\,dt \\ &=-i\Big(\exp(-\tfrac{t^3}{3})\exp(ix\omega t)|^\infty_0+\int^\infty_0\exp(-\tfrac{t^3}{3})t^2\exp(ix\omega t)\,dt\Big)\\ &=i -i\int^\infty_0\exp(-\tfrac{t^3}{3})t^2\exp(ix\omega t)\,dt \end{align} $$
Taking real parts and putting things together you obtain the result.
The integral representation given in the OP is \begin{align} Ai(x)&=\frac1\pi\int^\infty_0 \exp(-t^3/3)e^{-\frac12 xt}\mathfrak{R}\Big(e^{i(\frac{\pi}{6}+\frac{\sqrt{3}}{2}xt)}\Big)\,dt\\ &=\frac1\pi\int^\infty_0 \exp(-t^3/3)e^{-\frac12 xt}\,\cos\Big(\frac{\pi}{6}+\frac{\sqrt{3}}{2}xt\Big)\,dt \end{align}