- #1

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## Homework Statement

Solve [itex]x^2y'' - y = 0 [/itex] using Power Series Solution expanding about

*x*= 2.

_{o}## The Attempt at a Solution

First I expand the coefficient of y" (i.e. x

^{2}) about x

_{o}:

[tex]TS[x^2]|_{x_o=2} = 4+ 4(x - 2) + (x - 2)^2[/tex]

Assuming the solution takes the form:

[tex]y(x) = \sum_0^{\infty}a_n(x - 2)^n \Rightarrow y''(x) = \sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2}[/tex]

Plugging the above into the original DE gives:

[tex]

(4+ 4(x - 2) + (x - 2)^2)\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2} -

\sum_0^{\infty}a_n(x - 2)^n =

0

[/tex]

Distributing we have:

[tex]4\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2} +

4(x - 2)\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2} +

(x - 2)^2)\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2} -

\sum_0^{\infty}a_n(x - 2)^n =

0

[/tex]

[tex]\Rightarrow

4\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 2} +

4\sum_2^{\infty}n(n-1)a_n(x - 2)^{n - 1} +

\sum_2^{\infty}n(n-1)a_n(x - 2)^{n} -

\sum_0^{\infty}a_n(x - 2)^n =

0

\qquad(1)

[/tex]

Now the problem with (1) is twofold:

i) The indices are not the same and ii) the exponents on x are not the same.

I am not sure how to handle this. I was thinking that I could add 2 to the n index in the first term and add 1 to the second and then run the summations from zero. I think that this should work since the factors of 'n' 'n-1' in each term that I augment will cause each term to drop out at n = 0 and n = 1 and hence they will not contribute to the sum.

**Edit**: Fail. I failed to notice that when I add 2 to the dummy variable 'n' I will no longer have the factor of n to cause the term to drop out at n=0.

Any help on this?

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