- #1

Saladsamurai

- 3,020

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## Homework Statement

A Frobenius equation takes the form [itex] y'' + p(x)y' +q(x)y \qquad (1)[/itex] but for the sake of definiteness, let's take a particular example that my text uses:

[tex]6x^2y'' + 7xy' - (1 + x^2)y = 0 \qquad(2)[/tex]

We seek the solution in the form of [itex]y(x) = \sum_0^{\infty}x^{n+r}\qquad(3)[/itex]

The textbook says that after differentiating (3) and plugging back into (2) we arrive at:

[tex]

6x^2\sum_o^{\infty}(n+r)(n+r - 1)a_nx^{n+r - 2}

\,+\, 7x\sum_o^{\infty}(n+r)a_nx^{n+r-1}

\,-\, (1+x^2)\sum_o^{\infty}a_nx^{n+r}

\qquad(4)

[/tex]Ok great. Here is what I do not understand: The dummy indices. When we differentiate[itex]y(x) = \sum_0^{\infty}x^{n+r}\qquad(3)[/itex] don't we get [itex]\sum_1^{\infty}(n+r)a_nx^{n+r-1}\qquad(5)[/itex]? Note that the lower bound on the summation is now 1 not 0. Moreover, the lower bound on y''(x) should be 2 not 0.

Now I realize that when we do

*power series*solutions, we can just change the lower bounds back to 0, since there is always a factor that causes those first terms to vanish anyway. But here it does not appear that there are factors like that.

So why are the lower bounds on y' and y'' set back to 0?

Thanks