# Showing a series is a solution to a differential equation

I am attempting to show that the series $y(x)\sum_{n=0}^{\infty} a_{n}x^n$ is a solution to the differential equation $(1-x)^2y''-2y=0$ provided that $(n+2)a_{n+2}-2na_{n+1}+(n-2)a_n=0$

So i have: $$y=\sum_{n=0}^{\infty} a_{n}x^n$$ $$y'=\sum_{n=0}^{\infty}na_{n}x^{n-1}$$ $$y''=\sum_{n=0}^{\infty}a_{n}n(n-1)x^{n-2}$$

then substituting these into the differential equation I get:

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

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

relabeling the indexes:

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

and then cancelling the $n=-2$ and $n=-1$ terms:

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

but this doesn't give me what I want (I don't think) as I have $n^2$ terms as I would need

$(n^2+3n+2)a_{n+2}-(2n^2+n)a_{n+1}+(n^2-n-2)a_{n}=0$

I'm not sure where I have gone wrong?

Thanks very much for any help

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You are right till the last step.

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

which gives us $$\sum_{n=0}^{\infty} \left((n+2)(n+1) a_{n+2} - 2 n (n+1) a_{n+1} + (n^2-n-2)a_n \right)x^n = 0$$ and not $$\sum_{n=0}^{\infty} \left((n+2)(n+1) a_{n+2} - (2n^2+n) a_{n+1} + (n^2-n-2)a_n \right)x^n = 0$$ as you have written. Hence, setting the coefficients of $x^n$ to zero, we get that $$\left((n+2)(n+1) a_{n+2} - 2 n (n+1) a_{n+1} + (n^2-n-2)a_n \right)=0$$ Factorizing $n+1$ out, we get what you need i.e. $$\left((n+2)(n+1) a_{n+2} - 2 n (n+1) a_{n+1} + (n^2-n-2)a_n \right)= (n+1) \left((n+2) a_{n+2} - 2 n a_{n+1} + (n-2)a_n \right)$$ Hence, we get that $$(n+2) a_{n+2} - 2 n a_{n+1} + (n-2)a_n = 0$$

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Thanks very much! – hmmmm Apr 30 '12 at 9:30

Everything seems correct. Before you expand out powers of $n$, notice that equating like powers of $x^n$ in your last sum gives:

$(n+2)(n+1)a_{n+2}-2n(n+1)a_{n+1}+[n(n-1)-2]a_n=0$

Notice that $[n(n-1)-2]=n^2-n-2=(n+1)(n-2)$, so since $n\geq 0$, you can divide the recurrence equation by $(n+1)$ to get the desired result.

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You are correct.
Only you need to go on and observe that the lhs of your last equation factorizes as: $$(n+1)[(n+2)a_{n+2}-2n a_{n+1}+(n-2)a_n]$$

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