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

Question

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 $a_2=0$?

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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 $m>2$?

Answer 1

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.

Answer 2

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 $2a_2=0$.

Answer 3

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

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