Series Solutions to the Airy Differential Equation
Definition
The following differential equation is called the Airy differential equation.
$$ y^{\prime \prime}-xy=0,\quad -\infty<x<\infty $$
Explanation
The name originates from the British astronomer George Biddell Airy.
It is also called the Stokes equation.
Solution
Since the coefficient of $y^{\prime \prime}$ is $1$, all points are ordinary points. Among them, let’s find the power series solution around $x=0$. Assume that the solution of the Airy equation is as follows and converges in the interval $|x|<\rho$.
$$ y= \sum \limits _{n=0} ^{\infty} a_{n} x^n=a_{0}+a_{1}x+a_2x^2+\cdots $$
Then $y^{\prime \prime}$ is
$$ \begin{align*} y^{\prime \prime} =&\ \sum \limits_{n=2}^\infty n(n-1)a_{n}x^{n-2} \\ =&\ \sum \limits_{n=0}^\infty (n+2)(n+1)a_{n+2}x^n \\ =&\ 2\cdot 1 a_2+ 3\cdot2 a_{3}x +4\cdot 3 a_{4}x^2+\cdots \end{align*} $$
Substituting into the differential equation and matching the order of $x$, we get the following.
$$ \begin{align*} y^{\prime \prime}-xy =&\ \sum \limits_{n=0}^{\infty} (n+2)(n+1)a_{n+2}x^{n}-\sum \limits_{n=0}^{\infty}a_{n}x^{n+1} \\ =&\ \sum \limits_{n=-1}^{\infty} (n+3)(n+2)a_{n+3}x^{n+1}-\sum \limits_{n=0}^{\infty}a_{n}x^{n+1} \\ =&\ 2a_{2}+\sum \limits_{n=0}^{\infty} (n+3)(n+2)a_{n+3}x^{n+1}-\sum \limits_{n=0}^{\infty}a_{n}x^{n+1} \\ =&\ 2a_{2} + \sum \limits_{n=0}^{\infty } \left[ (n+3)(n+2)a_{n+3}-a_{n} \right]x^{n+1} \\ =&\ 0 \end{align*} $$
For any $x$ to always hold, all coefficients must be $0$. Therefore,
$$ a_{2}=0 $$
Arranging the recursion formula of the series coefficients for $a_{n+3}$, we get the following.
$$ a_{n+3}=\frac{a_{n}}{(n+3)(n+2)} $$
First, for $n=0$, we get the following.
$$ \begin{align*} a_{3} =&\ \frac{1}{3\cdot 2}a_{0} \\ a_{6} =&\ \frac{1}{6\cdot 5}a_{3} =\frac{1}{6\cdot 5 \cdot 3 \cdot 2}a_{0} \\ a_{9} =&\ \frac{1}{9\cdot 8}a_{6} =\frac{1}{9\cdot 8 \cdot 6\cdot 5 \cdot 3 \cdot 2}a_{0} \\ \vdots & \end{align*} $$
For $n=1$, we get the following.
$$ \begin{align*} a_{4} =&\ \frac{1}{4\cdot 3}a_{1} \\ a_{7} =&\ \frac{1}{7\cdot 6}a_{4} =\frac{1}{7\cdot 6 \cdot 4 \cdot 3}a_{1} \\ a_{10} =&\ \frac{1}{10\cdot 9}a_{7} =\frac{1}{10\cdot 9 \cdot 7\cdot 6 \cdot 4 \cdot 3}a_{1} \\ \vdots & \end{align*} $$
For $n=2$, we get the following.
$$ \begin{align*} a_{5} =&\ \frac{1}{5\cdot 4}a_{2}=0 \\ a_{8} =&\ \frac{1}{8\cdot 7}a_{5}=0 \\ a_{11} =&\ \frac{1}{11\cdot 10}a_{8} =0 \\ \vdots & \end{align*} $$
Therefore, the general solution of the Airy differential equation is as follows.
$$ \begin{align*} y =&\ \sum \limits_{n=0}^{\infty}a_{n}x^{n} \\ =&\ a_{0}+a_{1}x+a_{3}x^{3}+a_{4}x^{4}+a_{6}x^{6}+a_{7}x^{7}+\cdots \\ =&\ a_{0}+a_{1}x+\frac{1}{3\cdot 2}a_{0}x^{3}+\frac{1}{4\cdot 3}a_{1}x^{4}+\frac{1}{6\cdot 5 \cdot 3 \cdot 2}a_{0}x^{6}+\frac{1}{7\cdot 6 \cdot 4 \cdot 3}a_{1}x^{7} \\ =&\ a_{0}\left( 1+\frac{1}{3\cdot 2}x^{3}+\frac{1}{6\cdot 5 \cdot 3 \cdot 2}x^{6} + \cdots \right)+a_{1}\left( x+\frac{1}{4\cdot 3}x^{4}+\frac{1}{7\cdot 6\cdot 4\cdot 3}x^{7}+\cdots \right) \end{align*} $$
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