How to use Fourier Transform to solve the Airy's equation? 
Definition: If $f\in L^1(\mathbb{R}^n)$, the Fourier Transform of $f$ is the function $\hat{f}$ given by
  $$\hat{f}(y)=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}e^{-ix\cdot y}f(x)\;dx\;\;\;(y\in\mathbb{R}^n)$$
  and the Inverse Fourier Transform of $f$ is the function $f^\vee$ given by
  $$f^\vee(y)=\frac{1}{(2\pi)^{n/2}}\int_{\mathbb{R}^n}e^{ix\cdot y}\hat{f}(x)\;dx\;\;\;(y\in\mathbb{R}^n).$$

The problem is to use Fourier Transform to solve $(1)$.
$$\left\{\begin{matrix}
u_t(x,t)+u_{xxx}(x,t)=0, &(x,t)\in\mathbb{R}\times(0,\infty)  \\ 
 u(x,0)=g(x),& x\in\mathbb{R} 
\end{matrix}\right.\tag{1}$$
I found 
$$u(x,t)=\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}e^{iyx+iy^3t}\hat{g}(y)\;dy$$
Is it the correct answer? My calculation is below.
If we take the transform in $(1)$, we get
$$\left\{\begin{matrix}
\hat{u}_t(y,t)+(iy)^3\hat{u}(y,t)=0, &(y,t)\in\mathbb{R}\times(0,\infty)  \\ 
 \hat{u}(y,0)=\hat{g}(y),& y\in\mathbb{R} 
\end{matrix}\right.\tag{2}$$
If we solve the first equation in $(2)$ with respect $t$, we conclude that
$$\hat{u}(y,t)=\hat{g}(y)e^{iy^3t}$$
Therefore
$$u(y,t)=(\hat{u})^\vee(y,t)=\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}e^{ixy}\hat{u}(x,t)\;dx=\frac{1}{\sqrt{2\pi}}\int_\mathbb{R}e^{ixy}\hat{g}(x)e^{ix^3t}\;dx$$
 A: Disclaimer: I'm going to be working with a different definition of the Fourier transform.
$$\text{The Fourier transform } \mathcal{F}[f] \text{ is defined by } \hat{f}(k) = \int_{-\infty}^{\infty}e^{-ikx}f(x) dx.$$
$$\text{The inverse Fourier transform } \mathcal{F}^{-1}[f] \text{ is defined by } f(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty}e^{ikx}\hat{f}(k)dk$$
The convolution theorem and transform of the Airy function may require different scaling factors under your definition.
We proceed (as in your solution) to use the Fourier transform to reduce the PDE to an ODE, then solve this ODE to find
$$\hat{u}(y,t) = \hat{g}(y)\cdot e^{i y^{3}t}.$$
We want to get from $\hat{u}$ back to our original solution $u$. To do this we need the convolution theorem, which says that
$$ \mathcal{F} [ f(y) \ast g(y)] = \hat{f}(y) \cdot\hat{g}(y)$$
In our particular case, it tells us that 
$$u(x,t) = \mathcal{F}^{-1}[\hat{u}(k,t)] = \mathcal{F}^{-1}\left[\hat{g}(y)\cdot e^{i y^{3}t}\right] =g(x)\ast \mathcal{F}^{-1}\left[e^{i y^{3}t}\right].$$
We know (or can look up) that the Airy function transforms in the following way:
$$\mathcal{F}[\text{Ai}(x)] = e^{i\frac{k^{3}}{3}}.$$
We can use this to calculate the inverse transform that we need to find $u(x,t)$.
$$ \mathcal{F}^{-1}\left[e^{i y^{3}t}\right] = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{ikx}e^{i y^{3}t}dk $$
Make the change of variables $k = \kappa \frac{1}{\sqrt[3]{3t}}$. The integral becomes
$$\mathcal{F}^{-1}\left[e^{i y^{3}t}\right] = \frac{1}{2\pi} \frac{1}{\sqrt[3]{3t}} \int_{-\infty}^{\infty} e^{i\kappa(\frac{x}{\sqrt[3]{3t}})}e^{i\frac{\kappa^{3}}{3}}d\kappa = \frac{1}{\sqrt[3]{3t}} \text{Ai}\left(\frac{x}{\sqrt[3]{3t}}\right).$$
The final solution is then
$$u(x,t) = g(x)\ast \frac{1}{\sqrt[3]{3t}} \text{Ai}\left(\frac{x}{\sqrt[3]{3t}}\right).$$
