Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|cite|improve this question
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 $y=x^ku$ ,

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



share|cite|improve this answer

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.

share|cite|improve this answer

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 $p(x) = \sum_{n=0} p_n x^n$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.