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Given $(1+x^2)y''+2xy'-2y = 0$

The above equations obviously has analytic points everywhere except for $x=1$ and $-1$.

Find two linearly independent solutions $y_1$ and $y_2$ to the differential equation valid near $x_0=0$. To make life a little easier, choose the linearly independent equations:

$y_1$ $:$ $a_0$ = $y(x_0)$ = 1 and $a_1$ = $y'(x_0)$ = 0

$y_2$ $:$ $a_0$ = $y(x_0)$ = 0 and $a_1$ = $y'(x_0)$ = 1

After a mess of writing, I came up with the following:

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

I don't know if that monster is right, but that's where I need you help. Can somebody give this a sanity check, and then solve the rest?

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  • $\begingroup$ Some more detail would be nice. What did you get after substituting your power series ansatz into your differential equation? $\endgroup$ – J. M. is a poor mathematician May 7 '12 at 1:43
  • $\begingroup$ I checked the series solution myself, your expression of $a_{n+2}$ looks correct to me. $\endgroup$ – Shuhao Cao May 7 '12 at 2:14
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If we set up $y = \displaystyle \sum^{\infty}_{n=1} a_n x^n$ like you did, plugging back to the original equation: $$ (-2a_0+2a_2) + 6a_3 x +\sum^{\infty}_{n=2}\Big((n+2)(n+1)a_{n+2} - (n-1)(n+2)a_n\Big)x^n = 0 $$ simplify the expression for $a_{n+2}$ when $n\geq 2$: $$ a_{n+2} = -\frac{n-1}{n+1} a_n \qquad (*) $$ for the constant and $x$-term we have: $$ a_0 = a_2, \text{ and } a_3 = 0 $$

  • Now if the initial condition is $a_0 = 1, a_1 = 0$, then we have: $$ a_2 = 1, \text{ and } a_{n+2} = (-1)^{n/2}\frac{n-1}{n+1}\cdot \frac{n-3}{n-1}\cdots\frac{1}{3} = \frac{(-1)^{n/2}}{n+1} $$ and $n$ can be even numbers, let $n = 2k$ we have the solution is: $$ y = 1+x^2 + \sum^{\infty}_{k=1}\frac{(-1)^k x^{2k+2}}{2k+1} = 1 + x\cdot \sum^{\infty}_{k=0}\frac{(-1)^k x^{2k+1}}{2k+1} = 1+ x\arctan x $$

  • Now if the initial condition is $a_0 = 0, a_1 = 1$, $a_2 = a_0 = 0$ implies all even powered $x$ coefficients are zero after $n=2$ because of the relation $(*)$, also by $(*)$ and $a_3 = 0$ we know that all odd powered $x$ coefficients are zero too after $n=3$, therefore the solution is just: $$ y = x $$

To sum up, the two linearly independent solutions are: $$y = x \;\text{ or }\; 1+ x\arctan(x)$$

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  • $\begingroup$ Thanks Jon. The only reason why I got stuck was because the simplification threw me off. It was late and I didn't even think to simplify (in fact I thought it was already simplified). Anyway, thanks again. $\endgroup$ – Nico Bellic May 7 '12 at 14:16
  • $\begingroup$ @NicoBellic Haha, no problem, btw I love GTA4 too! $\endgroup$ – Shuhao Cao May 7 '12 at 17:01
  • $\begingroup$ Haha, nice few people recognize the reference. $\endgroup$ – Nico Bellic May 7 '12 at 18:16
  • $\begingroup$ The power series method works well provided I can solve the recursion relation. All the textbook examples usualy lead to a recursion that involves two coefficients only ( in this case $a_{n+2}$ and $a_n$). However, in most real world examples this is not the case, ie $a_{n+2}$ is usualy a function of $a_n$ and som other values of $a$ with lower indices. How do I solve the recursion relation then? Or in other words how would I go about solving an equation $(1 + x^2)y^{''} + 2 x y^{'} -2 x y =0$ using the power expansion method? $\endgroup$ – Przemo Feb 12 '14 at 16:27

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