# Finding a series expansion, about some given $x = x_0$ for the solution to a second order ODE

I am currently looking at a question requiring me to find a series expansion (about the point $x=0$) for the solution to the below second order ODE (Airy's equation):

$$y'' - xy = 0$$

I believe that I understand what this requires me to do, but I wish to make certain, so correct me if I am wrong:

By setting

$$y = \sum_{n=0}^{\infty} a_n x^n = a_0 + a_1x + a_2x^2 + a_3x^3 + ...$$

which also gives us

$$y'' = \sum_{n=2}^{\infty} n(n-1)a_nx^{n-2} = 2a_2 + 6a_3x + 12a_4x^2 + 20a_5x^3 + ...$$

and subbing these power series into the ODE gives us

$$2a_2 + 6a_3x + 12a_4x^2 + 20a_5x^3 + 30a_6x^4 + ... - a_0x - a_1x^2 - a_2x^3 - a_3x^4 + ... = 0$$

And thus, setting $a_0 = a_0$ and $a_1 = a_1$ arbitrarily, we can express every other $a_n$ in terms of $a_0, a_1$, as follows:

$$2a_2 = 0 \Rightarrow a_2 = 0 \\ 6a_3 - a_0 = 0 \Rightarrow a_3 = \frac{a_0}{6} \\ 12a_4 - a_1 = 0 \Rightarrow a_4 = \frac{a_1}{12} \\ 20a_5 - a_2 = 0 \Rightarrow a_5 = \frac{a_2}{20} = 0\\ 30a_6 - a_3 = 0 \Rightarrow a_6 = \frac{a_3}{30} = \frac{a_0}{180}$$

And thus we can express the solution of the ODE in the form

$$y = a_0 \left( x + \frac{x}{6} + \frac{x}{180} + \dots \right) + a_1 \left( x + \frac{x}{12} + \dots \right)$$

Am I correct in thinking that these are our series solutions, and the correct answer to the question?

If so, where have I satisfied the fact that the expansion should be about $x=0$?

Lastly, the question goes on to say "write the explicit form of the solutions $y_1, y_2$ such that the general solution of the ODE is $y = a_0 y_1 + a_1 y_2$. Is this not what we have already done in the above?

• You assumed $y =\sum a_n(x-x_0)^n$, in which $x_0=0$, hence it is the taylor expansion evaluated at $x_0=0$. – MrYouMath Feb 6 '17 at 15:43