# How to have the right number of solutions when recurrence relation for second order ODE gives relation between coefficients $a_n$ & $a_m$, $m>2$?

I have an equation: $$y''-xy=0$$ and I am told to find two linearly independent power series solutions, about x = 0. I am looking for solutions in the form $$\sum_0^{\infty}a_nx^n$$ and successfully obtained the recurrence relation: $$(n+3)(n+2)a_{n+3}=a_n$$ Which makes me think that I can freely choose $$a_0$$, $$a_1$$ & $$a_2$$. Problem: that is 3 linearly independent solultion, and for a second order ODE I am supposed to have max $$2$$.

I am told that $$a_2=0$$. This would solve the problem about number of solutions. How can I conclude that $$a_2=0$$?

Question phrased in a more general way: how to come up with only $$2$$ solutions for second order ODE if recurrence realtion is between $$a_n$$ & $$a_m$$, where $$m>2$$?

With the convention $$a_n=0$$ for $$n<0$$, the first 3 non-trivial equations read \begin{align} n&=-3:& 0a_0&=a_{-3}=0\\ n&=-2:& 0a_1&=a_{-2}=0\\ n&=-1:& 2a_2&=a_{-1}=0 \end{align} so that indeed $$a_0,a_1$$ are free, while the value of $$a_2$$ is fixed to zero.

When substituting $$y = \sum_{k=0}^{\infty}a_k x^k$$ into the DE, as

$$\frac{d^2}{dx^2}\left(\sum_{k=0}^{\infty}a_k x^k\right)-x \sum_{k=0}^{\infty}a_k x^k=0$$

the resulting "polynomial" should be identically null, or

$$2a_2 -(a_0-6a_3)x - (a_1-12a_4)x^2+\cdots + = 0$$

from this condition we can establish the recurrence, and also $$2a_2=0$$.

• "polynomial"?... – lhf Jan 8 at 13:20
• Thanks for the hint. – Cesareo Jan 8 at 13:37

$$\begin{array}{r} y''&= &2a_2 &+& 6 a_3 x &+& 12 a_4 x^2 &+& 20 a_5 x^3 &+& \cdots \\ xy &= &0 &+& a_0 x &+& a_1 x^2 &+& a_2 x^3 &+& \cdots \end{array}$$ implies $$2a_2=0$$.

Alternatively, and much simpler, evaluating $$y''(x)-xy(x)=0$$ at $$x=0$$ gives $$y''(0)=0$$.