Airy differential equation $y''+(x-\frac14)y=0$ $y''+(x-\frac14)y=0$
this is the equation I am trying to solve
mathematica tells me the answer is:
$$\left.c_1 \text{Ai}\left(-\sqrt[3]{-1} \left(\frac{1}{4}-x\right)\right)+c_2 \text{Bi}\left(-\sqrt[3]{-1} \left(\frac{1}{4}-x\right)\right)\right]$$
how do I get this result. I don't know how I would end up with $\sqrt[3]{-1}$ in the argument
 A: $Ai(x)$ and $Bi(x)$ are solutions to the differential equation:
$\frac{d^2y}{dx^2}-xy=0$
Obviously, this is different from the form given above.
Thus, let $(x-\frac{1}{4})=-t$
We now have,
$\frac{d^2y}{dx^2}-ty=0$
We have, $\frac{dy}{dx}=\frac{dy}{dt}\cdot\frac{dt}{dx}=-\frac{dy}{dt}$
Thus, $\frac{d^2y}{dx^2}=-\frac{d^2y}{dt^2}\cdot -1=\frac{d^2y}{dt^2}$
We now have, 
$\frac{d^2y}{dt^2}-ty=0$
This is exactly the needed form.
Writing the solution as a function of $t$, we have
$y=c_1Ai(t)+c_2Bi(t)$
But now, setting $t$ in terms of $x$, we have
$y=c_1Ai(\frac{1}{4}-x)+c_2Bi(\frac{1}{4}-x)$
As you can also see, $-\sqrt[3]{-1}=1$ which is the same as the above.
As for Mathematica bringing it up in such a form, I believe it is due to the manipulations given above. When it set $t$ in terms of $x$ and ended up calculating the first and second derivatives in terms of $t$, then set $t$ in terms of $x$ back in the solution, the $-\sqrt[3]{-1}$ came up. You do not have to be bothered with it. It's because of its mechanical nature.
A: The Airy functions $\operatorname{Ai}$ and $\operatorname{Bi}$ are two fundamental solutions of the differential equation:
$$ y'' = x y \tag{1}$$
and they can be represented by the integrals:
$$ \operatorname{Ai}(x)=\frac{1}{\pi}\int_{0}^{+\infty}\cos\left(\frac{t^3}{3}+xt\right)\,dx, $$
$$ \operatorname{Bi}(x)=\frac{1}{\pi}\int_{0}^{+\infty}\exp\left(-\frac{t^3}{3}+xt\right)+\sin\left(\frac{t^3}{3}+xt\right)\,dx, \tag{2}$$
so they are entire functions on the complex plane having non-integer order, $\frac{3}{2}$. 
The solutions of your differential equation can be derived from $(1)$ with the change of variable $\left(x-\frac{1}{4}\right)=\omega z$ where $\omega$ is a primitive third root of unity.
