# Power series solution for $y''-xy'-y =0$

I must find a power series solution for:

$$y''-xy'-y =0$$

centered at $x_0 = 0$

I'm supposing it in the form:

$$y = \sum_{n=0}^{\infty}a_nx^n$$

so:

$$y' = \sum_{n=1}^{\infty}na_nx^{n-1}$$ $$y'' = \sum_{n=2}^{\infty}n(n-1)a_nx^{n-2}$$

I've shifted the index of all of them, so I got:

$$y = \sum_{n=0}^{\infty}a_nx^n$$ $$y' = \sum_{n=0}^{\infty}(n+1)a_{n+1}x^{n}$$ $$y'' = \sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}$$

so,substituting it in the differential equation, I get:

$$\sum_{n=0}^{\infty}(n+2)(n+1)a_{n+2}x^{n}-x\sum_{n=0}^{\infty}(n+1)a_{n+1}x^{n}-\sum_{n=0}^{\infty}a_nx^n = 0$$

The idea now is to group them all and isolate the $x^n$ coefficient, but what should I do with that $x$ multiplying $y'$? It'll shift the index of the sum and it'll have the term $x^{n+1}$ in it, making it impossible to isolate $x^n$

• So isolate $x^{n+1}$. You might need to consider the first few values $a_0, a_1,..$ separately as they are not present in all three sums. – Paul Nov 25 '16 at 11:44
• It's probably more efficient if you multiply the $x$ inside the sum before you shift the index. – B. Goddard Nov 25 '16 at 11:45
• @B.Goddard but I have two choices: either I start at $n=0$ and $x^{n+1}$ or I start at $n=1$ and $x^n$, I can't do both :c – Guerlando OCs Nov 25 '16 at 15:07
• The standard way is to multiply the $x$ inside. Then shift the index, so you end up starting at $n=1$. Then take the $n=0$ terms out of the other two sums and move them to the front of the expression. Now all three sums start at $n=1$. – B. Goddard Nov 25 '16 at 15:09
• Put the x in the middle term inside the summation and shift the index again.. – DanielWainfleet Oct 18 '17 at 20:00

If $y=\sum_{n\ge0}a_nx^n$, then $$y'=\sum_{n\ge1}na_nx^{n-1}$$ and $$y''=\sum_{n\ge2}n(n-1)a_nx^{n-2}=\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n$$ Therefore $$y''-xy'-y=\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n-\sum_{n\ge0}na_nx^{n}-\sum_{n\ge0}a_nx^n$$ so $$(n+2)(n+1)a_{n+2}-na_n-a_n=0$$ or $$(n+2)a_{n+2}-a_n=0$$