# Derivatives of a Series

## Homework Statement

This is for a differential equations class I'm taking and we're talking about the method of Frobeneus, Euler equations, and power series solutions for non-constant coefficients. The ODE is:
$$6x^2y''+7xy'-(1-x^2)y=0$$

I need to find the recurrence formula and I keep running into a problem with my grouping. So the question I have is when I take the derivative of a sum, does that change its index, or does the index stay the same? Because when the book explains it with power series the index is changed with each derivative taken. But when they do the Frobeneus method the index does not change when a derivative is taken.

## The Attempt at a Solution

Assume $$y=\sum_{n=0}^{\infty}a_nx^{n+r}$$ where r is the root of the indicial and then the derivatives are:
$$y'=\sum_{n=1}^{\infty}(n+r)a_nx^{n+r-1}$$ and $$y''=\sum_{n=2}^{\infty}(n+r-1)(n+r)a_nx^{n+r-2}$$
Plugging this into the ODE gives:
$$6\sum_{n=2}^{\infty}(n+r-1)(n+r)a_nx^{n+r} +7\sum_{n=1}^{\infty}(n+r)a_nx^{n+r}-\sum_{n=0}^{\infty}a_nx^{n+r}+\sum_{n=0}^{\infty}a_nx^{n+r+2}$$

Here is where the problem begins, I've tried changing my indices, but I can't quite get the powers of x and the indices to agree, and when I do get the indices to agree I get three different values of a ($a_1$,$a_2$, and $a_0$)

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## Homework Statement

This is for a differential equations class I'm taking and we're talking about the method of Frobeneus, Euler equations, and power series solutions for non-constant coefficients. The ODE is:
$$6x^2y''+7xy'-(1-x^2)y=0$$

I need to find the recurrence formula and I keep running into a problem with my grouping. So the question I have is when I take the derivative of a sum, does that change its index, or does the index stay the same? Because when the book explains it with power series the index is changed with each derivative taken. But when they do the Frobeneus method the index does not change when a derivative is taken.

## The Attempt at a Solution

Assume $$y=\sum_{n=0}^{\infty}a_nx^{n+r}$$ where r is the root of the indicial and then the derivatives are:
$$y'=\sum_{n=1}^{\infty}(n+r)a_nx^{n+r-1}$$ and $$y''=\sum_{n=2}^{\infty}(n+r-1)(n+r)a_nx^{n+r-2}$$
Plugging this into the ODE gives:
$$6\sum_{n=2}^{\infty}(n+r-1)(n+r)a_nx^{n+r} +7\sum_{n=1}^{\infty}(n+r)a_nx^{n+r}-\sum_{n=0}^{\infty}a_nx^{n+r}+\sum_{n=0}^{\infty}a_nx^{n+r+2}$$

Here is where the problem begins, I've tried changing my indices, but I can't quite get the powers of x and the indices to agree, and when I do get the indices to agree I get three different values of a ($a_1$,$a_2$, and $a_0$)
In $y$ the successive powers are $r, r+1, r+2, r+3, \ldots$, while in $y'$ they are $r-1, r , r+1, r+2 , \ldots$, so for the same power of $x$ you need to shift the index on the coefficient by 1; that is, from $a_j x^{r+j}$ in $y$ you get $(r+j) a_j x^{r+j-1}$ in $y'$. If we let $n = r+j$ then (with $b_{r+j} = a_j$) we have $a_n x^n$ in $y$ but $n a_n x^{n-1}$ in $y'$, and that last one can be re-written as $(m+1) a_{m+1} x^m$, where $m = n-1$. You can choose to write things however you want, as long as you are careful and avoid making mistakes.