I am looking for help on the following, Show that the for odd non-negative integers $m$ there are a polynomials of degree $m$ when $r=1$? This is from the hermite differential equation, have showed the recurrence relation and the indicial equation implies $r=0$ and $r=1$, when the case of $r=1$
I am aware that $a_0$ is free and $a_1=0$, however not sure how to go about that there are polynomials of degree $m$ when $r=1$?
My Hermite equation is y''+2xy'+2my=0
My recurrance relation is $a_{n+2}=\frac{2(n+r-m)a_n}{(n+r+2)(n+r+1)} n=0,1...$
My indicial equation is the lowest order term:indicial being $(r)(r-1){a_0}{x^{-2+r}}+(1+r)(r){a_1}{x^{-1+r}}=0$ hence
$(r)(r-1))=0$ hence r=0 and r=1, with $y(x)=\sum_{n=0}^{\infty}{a_n}{x^{n+r}}$ and Many thanks in advance.