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I try to solve this differential equation whose solution seems not to be constructable in power series $y''+(x+a/x^2+b)y=0$, where $a$ and $b$ are some positive real numbers. If one can help me please?

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Is the coefficient of $y$ supposed to be $x+\frac{a}{x^2} + b$, or $\frac{x+a}{x^2+b}$? The slash notation is ambiguous; please avoid it. – Arturo Magidin Feb 3 '12 at 20:27

3 Answers 3

Let us solve this problem in detail to show what kind of problems we encounter in here. We insert the ansatz $y(x) = \sum\limits_{n=0}^\infty p_n x^{n+\alpha}$ into the ODE and then equate to zero the coefficients at consecutive powers. We have: \begin{eqnarray} coeff @ x^{\alpha-2} :&& p_0 \alpha (\alpha-1) + a p_0 =0 \\ coeff @ x^{\alpha-1}:&& p_1 (\alpha+1) \alpha + a p_1 = 0 \\ coeff @x^{\alpha} :&& p_2(\alpha+2)(\alpha+1) + a p_2 + b p_0 = 0 \end{eqnarray} This gives $\alpha = \frac{1}{2} \left( 1 \pm \sqrt{1- 4 a} \right)$ and $p_1=0$ and $p_2= -b p_0/(2 (2 \pm \sqrt{1- 4 a}))$. The recursion relation for the coefficients now reads: \begin{equation} p_{n+2} \left( (n+2+\alpha)(n+1+\alpha) + a \right) + p_n b + p_{n-1}=0 \end{equation} for $n=1,2,\dots$. Now we substitute $p_n \rightarrow p_{n+1}$ and we get a following recursion relation: \begin{equation} p_{n+3} = f_n \left( b \cdot p_{n+1} + p_n \right) \end{equation} for $n=1,2,3,\dots$ with $f_n := - ((n+2)(n+2 \pm\sqrt{1-4 a}))^{-1}$ and subject to $(p_1,p_2)= (1,0)$. By looking at this recursion relation we see that the solution is some polynomial in the variable $b$. Is it possible to find a closed form expression for all coefficients of that polynomial? By starting from a given value of $n$ we can back-propagate this equation following two rules, firstly only steps of length two or three are allowed, secondly a step by two and three is assigned a factor $b$ and a unity respectively. We denote by $(i_l + l-1)$ the positions of the $f_{n-2(i_l-1)-3 \cdot l}$-factors related to movements of length three. Here for $l=1,\dots,s$. All the remaining $f$-factors are related to steps of length two. Since at the end of the back-propagation process we must always hit unity this yields a following constraint: $n - 2(i_{s+1}-1) - 3 s= 1$. Due to the initial conditions he exponent assigned to all the terms in question equals $i_{s+1}-1$. If we now consider two cases of $n$ being even and odd respectively then we easily arrive at the following result: Let $n \notin 2 {\mathbb N}$, ie $n=2 m+1$. Then we have: \begin{eqnarray} p_{n+3} &=& \sum\limits_{j=0}^{\lfloor \frac{m}{3} \rfloor} b^{m- 3 j} \cdot \sum\limits_{1 \le i_1 \le i_2 \le \dots \le i_{2 j} \le m+1- 3j} \left( \prod\limits_{l=1}^{2 j+1} \prod\limits_{\xi=0}^{i_l-i_{l-1}-1} f_{n-2 \cdot i_{l-1} - 3 (l-1) - 2 \xi}\right) \cdot \left(\prod\limits_{l=1}^{2 j} f_{n-2 \cdot (i_l-1)- 3 l} \right)\\ &=& {\mathcal C}_m \sum\limits_{j=0}^{\lfloor \frac{m}{3} \rfloor} b^{m- 3 j} \frac{(-1)^{1-j+m}}{2^{2(1+j+m)}} \pi^{2 j} \sum\limits_{1 < i_1 < i_2 < \dots < i_{2 j} \le m+1- j} \prod\limits_{l=1}^{2 j} \prod\limits_{p=\pm} \binom{\frac{\theta_p-n}{2} + \frac{1}{2} l -2 + i_l}{\frac{1}{2}} \end{eqnarray} where \begin{equation} {\mathcal C}_m := \prod\limits_{p=\pm} \frac{(\frac{\theta_p-n}{2}-1)!}{(\frac{\theta_p-n}{2}+m)!} \end{equation} Likewise let $n \in 2{\mathbb N}$, ie $n=2 m$. Then we have: \begin{eqnarray} p_{n+3} &=& \sum\limits_{j=0}^{\lfloor \frac{(m-2)}{3} \rfloor} b^{m- 3 j-2} \cdot \sum\limits_{1 \le i_1 \le i_2 \le \dots \le i_{2 j+1} \le m-1- 3 j} \left( \prod\limits_{l=1}^{2 j+2} \prod\limits_{\xi=0}^{i_l-i_{l-1}-1} f_{n-2 \cdot i_{l-1} - 3 (l-1) - 2 \xi}\right) \cdot \left(\prod\limits_{l=1}^{2 j+1} f_{n-2 \cdot (i_l-1)- 3 l} \right)\\ &=&{\mathcal D}_m \sum\limits_{j=0}^{\lfloor \frac{(m-2)}{3} \rfloor} b^{m- 3 j-2} \frac{(-1)^{m-j}}{2^{2(1+j+m)}} \pi^{1+2 j} \sum\limits_{1 < i_1 < i_2 < \dots < i_{2 j+1} \le m- j} \prod\limits_{l=1}^{2 j+1} \prod\limits_{p=\pm} \binom{\frac{\theta_p-n}{2} + \frac{1}{2} l -2 + i_l}{\frac{1}{2}} \end{eqnarray} where \begin{equation} {\mathcal D}_m := \prod\limits_{p=\pm} \frac{(\frac{\theta_p-n}{2}-1)!}{(\frac{\theta_p-n}{2}+m-\frac{1}{2})!} \end{equation} and $\theta_\pm = (-2,-2 \pm \sqrt{1-4 a})$. We wrote a simple Mathematica program to verify that those results are correct. Now comes the most interesting part. Is it possible to do the sums over the $i$-indices analytically or otherwise the results above are only some sort of perturbation results in the variable $b$. We leave this question open for the time being.

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The differential equation $$y''+ \left(x+ \frac{a}{x^2}+b \right)y=0$$ has a regular singular point at $x=0$. In such a case, it is not always possible to construct a power series solution. However, it is always possible to find a solution of the form $$y = x^\alpha p(x)$$ with $\alpha \in \mathbb{C}$ and $p(x) = \sum_{n=0} p_n x^n$.

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And what if OP really meant $$y''+{x+a\over x^2+b}y=0$$ – Gerry Myerson Mar 4 '12 at 23:51
@Gerry: then he should be able to develop $y(x)$ in a power series around $x=0$. – Fabian Mar 5 '12 at 3:11




Let $y=x^ku$ ,

Then $y'=x^ku'+kx^{k-1}u$


$\therefore x^2(x^ku''+2kx^{k-1}u'+k(k-1)x^{k-2}u)+(x^3+bx^2+a)x^ku=0$



Choose $k(k-1)+a=0$ , i.e. $k=\dfrac{1\pm\sqrt{1-4a}}{2}$ , the ODE becomes



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