Let u(x)= \sum_{n=0}^\infty a_nx^n[/tex]. Then u&#039;= \sum_{n=1}^\infty na_nx^{n-1} and u&#039;&#039;= \sum_{n=2}^\infty n(n-1)a_nx^{n-2}.<br />
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(I have changed the beginning index from 0 to 1 to 2 because each term, in the sum for u', is multiplied by n and so is 0 when n= 0 and, in the sum for u'' is multiplied by n(n- 1) and so is 0 when n= 0 and when n= 1.)<br />
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Replacing u'' and u in your equation with those, <br />
\sum_{n= 2}^\infty n(n-1)a_nx^{n-2}+ x^2\sum_{n=0}^\infty a_nx^n= 0<br />
\sum_{n=2}^\infty n(n-1)a_nx^{n-2}+ \sum_{n=0}^\infty a_nx^{n+2}= 0<br />
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In order to combine "like powers", let i= n-2 in the first sum. Then n= i+ 2, when n=2, i= 0 so the sum is<br />
\sum_{i=0}^\infty (i+2)(i+1)a_{i+2}x^i.<br />
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Let i= n+ 2 in the second sum (since the indices have meaning only in the individual sums, we can do this.). Now n= i- 2. When n= 0, i= 2 so the sum becomes<br />
\sum_{i= 2}^\infty a_{i-2}x^i<br />
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The differential equation becomes <br />
\sum_{i=0}^\infty (i+2)(i+1)a_{i+2}x^i+ \sum_{i=2}^\infty a_{i-2}x^i<br />
and can compare "like powers" by looking at specific values of i. <br />
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The second sum does not start until i= 2 so for i= 0 and i= 1 we have<br />
(0+2)(0+ 1)a_2= 2a_2= 0 which tells us that a_2= 0 and (1+ 2)(1+ 1)a_3= 6a_3= 0 which tells us that a_3= 0.<br />
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For i greater than 1, we have <br />
(i+ 2)(i+ 1)a_{i+2}+ a_{i-2}= 0<br />
or<br />
a_{i+2}= -\frac{a_{i-2}}{(i+ 2)(i+1)}<br />
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That gives a recursive equation for the coefficients. Since "n-2" will not be non-negative until n= 2, where a+ 2= 4, we cannot, of course, determine a_0, a_1, a_2, or a_3 from this, but that's okay- we already know that a_2= a_3= 0 and we expect a second order d.e. to involve two "constants". We can take those constants to be a_0 and a_1. (In fact, it is easy to see that y(0)= a_0 and y&#039;(0)= a_1, "initial conditions" for this problem. If you were given other initial conditions, say y(x_0) and y&#039;(x_0), write the power series in terms of (x- x_0)^n.)<br />
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Now, we have, with i= 2, <br />
a_4= \frac{a_0}{(2+ 2)(2+1)}= \frac{a_0}{8}<br />
with i= 3,<br />
a_5= \frac{a_1}{(3+2)(3+ 1)}= \frac{a_1}{18}<br />
with i= 4,<br />
a_6= \frac{a_2}{(4+ 2)(4+ 1)}= 0<br />
because a_2= 0.<br />
Similarly, with i= 5, <br />
a_7= \frac{a_3}{(5+2)(5+1)}= 0<br />
because a_3= 0<br />
With i= 6,<br />
a_8= \frac{a_4}{(6+2)(6+1)}= \frac{a_4}{56}<br />
= \frac{a_0}{8(56)}= \frac{a_0}{448}<br />
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etc.<br />
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It might be best to look at the indices "modulo 4". If i is 3 mod 4 (3, 7, 11, 15,...) a_i= 0. If i is 2 mod 4 (2, 6, 10, 14, ...), a_i= 0.<br />
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If i is 0 mod 4 (0, 4, 8, 12, ...) a_i will be a_0 divided by an integer. We could factor a_0 out of that sum and have a_0 times some function. If i is 1 mod 4 (1, 5, 9, 13, ...) a_i[/tex] will be a_1 divided by some integer. We could factor a_1 out of that sum and have a_1 times some function. The result would be <br />
u(x)= a_0f(x)+ a_1g(x)<br />
where f and g are solutions to the differential equation. Of course f and g may be extremely complicated sums. You cannot, in general, expect to be able to determine a formula for the sums or even for the general coefficients.
