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I'm attempting to find a recurrence relation and the first few terms of a power series solution for the differential equation:

$$(1-x^2)y \prime\prime - 2xy\prime + \lambda y = 0$$

Where $\lambda$ is some integer.

So I've assumed a solution $y = \sum_{n=0}^\infty a_nx^n$, and so forth for $y\prime$ and $y\prime\prime$.

However, whenever I plug in and try to derive the recurrence relation, I'm not able to reduce it nicely, and as such I can't figure the first few terms of the series.

I'm sort of new to differential equations in general so please bear with my naiveté.

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For the differential equation $(1-x^{2}) y'' - 2 x y' + \lambda y = 0$ it is seen that if $y$ is of the form $\sum a_{n} x^{n}$ the following holds. \begin{align} y(x) &= \sum_{n=0}^{\infty} a_{n} \, x^{n} \\ y'(x) &= \sum_{n=0}^{\infty} n \, a_{n} \, x^{n-1} \\ y''(x) &= \sum_{n=0}^{\infty} n(n-1) \, a_{n} \, x^{n-2} \end{align} \begin{align} 0 &= (1 - x^{2} ) \, \sum_{n=0}^{\infty} n(n-1) \, a_{n} x^{n-2} - 2 x \, \sum_{n=0}^{\infty} n \, a_{n} \, x^{n-1} + \lambda \sum_{n=0}^{\infty} a_{n} \, x^{n} \\ &= \sum_{n=2}^{\infty} n(n-1) \, a_{n} \, x^{n-2} - \sum_{n=0}^{\infty} (n(n+1) - \lambda) \, a_{n} \, x^{n} \\ &= \sum_{n=0}^{\infty} (n+2)(n+1) \, a_{n+2} \, x^{n} - \sum_{n=0}^{\infty} (n(n+1) - \lambda) \, a_{n} \, x^{n} \\ &= \sum_{n=0}^{\infty} \left[ (n+2)(n+1) \, a_{n+2} - (n(n+1) - \lambda) \, a_{n} \right] \, x^{n} \end{align} From this equation the coefficient equation can be obtained. It is \begin{align} a_{n+2} = \frac{n(n+1) - \lambda}{(n+1)(n+2)} \, a_{n} \hspace{10mm} n \geq 0. \end{align}

Now that the recurrence relation has been obtained. Try a few values of $n$ to obtain the first few terms. The first two terms are defined as $a_{0}, a_{1}$ and the remaining are to follow. \begin{align} a_{2} &= \frac{- \lambda }{2!} \, a_{0} \\ a_{3} &= \frac{2-\lambda}{2 \cdot 3} \, a_{1} = \frac{(-1) (\lambda - 2)}{3!} \, a_{1} \\ a_{4} &= \frac{6 - \lambda}{3 \cdot 4} \, a_{2} = \frac{(-1)^{2} \lambda (\lambda - 6)}{4!} \, a_{0} \end{align} and so on. The solution for $y(x)$ is of the form \begin{align} y(x) &= a_{0} \left[1 - \frac{\lambda}{2!} \, x^{2} + \frac{(-1)^{2} \lambda (\lambda -6)}{4!} \, x^{4} + \frac{(-1)^{3} \lambda (\lambda -6)(\lambda -20)}{6!} \, x^{6} + \cdots \right] \\ & \hspace{5mm} + a_{1} \left[x + \frac{(-1)(\lambda - 2)}{ 3!} \, x^{3} + \frac{(-1)^{2} (\lambda - 2)(\lambda - 12)}{5!} \, x^{5} + \cdots \right] \end{align}

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  • $\begingroup$ So this is the path I took towards determining the recurrence relation as well. This may be a stupid question, but how do I use this relationship to determine the first few terms of the power series itself? $\endgroup$ – uknowbighank Mar 30 '15 at 5:51
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    $\begingroup$ @uknowbighank. Since the equation is second order, you have two arbitrary coefficients, say $a_0$ and $a_1$. These could be fixed by conditions. $\endgroup$ – Claude Leibovici Mar 30 '15 at 5:55
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    $\begingroup$ @uknowbighank The statement made by Claude Leibovici is correct and to demonstrate how it is applied the resulting series has been shown. $\endgroup$ – Leucippus Mar 30 '15 at 6:05
  • $\begingroup$ Okay, the problem didn't stipulate any conditions, so I suppose I'm to treat $a_0$ and $a_1$ as arbitrary constants. But your answer has greatly clarified how I should approach this sort of problem, so thank you. $\endgroup$ – uknowbighank Mar 30 '15 at 6:09
  • $\begingroup$ @Leucippus: Can we perform such calculations for recurrence relations using Maple. $\endgroup$ – IgotiT Nov 24 '16 at 0:47
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You have $$y = \sum_{n=0}^\infty a_nx^n$$ $$y' = \sum_{n=0}^\infty n a_nx^{n-1}$$ $$y'' = \sum_{n=0}^\infty n(n-1) a_nx^{n-2}$$ $$(1-x^2)y'' - 2xy' + \lambda y =y'' -x^2y''-2xy' + \lambda y =0$$ Now, replace $y,y',y''$by their expression for the last right hand side. So, $$\sum_{n=0}^\infty n(n-1) a_nx^{n-2}-\sum_{n=0}^\infty n(n-1) a_nx^{n}-2\sum_{n=0}^\infty n a_nx^{n}+\lambda\sum_{n=0}^\infty a_nx^n=0$$ Now, write the term for a given power $m$ of $x$. This will give $$(m+2)(m+1)a_{m+2}-m(m-1)a_m-2ma_m+\lambda a_m=0$$

I am sure that you can take from here.

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