For each $a \in \mathbb{Z}^+$ let the following ODE

$$ x'' - \dfrac{a (a+1)}{(1 +t^2)} x = 0$$

  • Using power series around the origin, show that the equation has a solution $p_a(t)$ which is a polinomial with degree $a+1$
  • Using an appropiate reduction order method, show that the general solution to the ODE is $$x = p_a(t) \left( k_1 + k_+2 \int \dfrac{dt}{p_a^2(t)} \right)$$

My attempt

Rewriting the equation as

$$(1+t^2) x'' - a (a+1)x = 0$$

Suppose a power series near 0 solution, then

$$x(t) = \sum_{k=0}^\infty a_l t^k \qquad x'(t) = \sum_{k=1}^\infty k \, a_k t^{k+-1} \qquad x''(t) = \sum_{k=2}^\infty k(k-1) a_k t^{k-2}$$

Introducing these expresions in the ODE we get

$$ (1 +t^2) \sum_{k=2}^\infty k(k-1) a_k t^{k-2} - a (a+1) \sum_{k=0}^\infty a_k t^k = 0 \implies$$ $$ \implies \sum_{k=0}^\infty (k+2)(k+1) a_{k+2} t^k + \sum_{k=2} ^\infty k(k-1) a_k t^k -a(a+1) \sum_{k=0}^\infty a_k t^k = 0$$

Now we should solve the following recurrence relation

$$\begin{cases} k=0 \qquad \qquad 2 a_2 - a(a+1) a_0 \\ k=1 \qquad \qquad 6 a_3 - a(a+1) a_1 = 0 \\ k \ge2 \qquad \qquad (k+2)(k+1)a_{k+2} + k(k-1) a_k - a (a+1) a_k =0 \end{cases}$$

I have solved this last recurrence relation with Mathematica using RSolve but the solution is $a_k = 0 \, \forall k$

Where am I wrong?

  • $\begingroup$ Why don't you keep your sum from $k=0$. Your last formula applies for all $k$'s, including $0$ and $1$. $\endgroup$ Jan 28, 2014 at 10:58
  • $\begingroup$ You're right, but still how can I solve that recurrence relation? Is the solution given by mathematica the only solution? $\endgroup$
    – José D.
    Jan 28, 2014 at 11:05

3 Answers 3


I think you are not using Fröbenius method correctly, shouldn't the solution be expressed as it follows:

$$x(t) = \sum^\infty_{l=0} b_l t^{l+s}?$$

where $s$ is to be solved from the indicial equation, which is $s (s-1) = 0$ and $b_l$ can be obtained by the recurrence relation. If you give me some minutes, I can try to solve your question.


Edit: if I am not mistaken, which is perfectly possible, and you let me change your notation, the Fröbenius method leads to the equation:

$$\sum^\infty_{n=0} b_n t^{n+s-2}(n+s-1)(n+s) + \sum^\infty_{n=0} \left[(n+s)(n+s-1)-(a+1)a \right] t^{n+s} = 0,$$

which can be rewritten as follows:

$$\sum^\infty_{n=-2} b_{n+2} t^{n+s}(n+s+2)(n+s+1) + \sum^\infty_{n=0} \left[(n+s)(n+s-1)-(a+1)a \right] t^{n+s} = 0,$$

and, as the first sum contains two more terms, it yields to the indicial equation:

$$\begin{align} n = -2: & \quad b_0 t^{s-2} s (s-1) = 0 \Rightarrow s = 1 \ \text{or } s = 0,\\ n = -1: & \quad b_1 t^{s-1} (s+1)(s) = 0, \end{align} $$ since the roots of the indicial equation differ by an integer there may be pathology and one of the index may not be used in order to get the two linearly indepent parts of the solution. If you choose $s=1$ (as my teacher says, in order to avoid problems), then the recurrence relation holds:

$$b_{n+2} = \frac{a(a+1)-(n+1)}{(n+3)(n+2)}b_n, \quad n \geq 0$$

but you can easily prove that $b_{2p+1} = 0$, so every odd term vanishes. If you compute the values $b_2, b_4, \ldots$ in terms of $b_0$ you will have (Fröbenius):

$$y_1(t) = t^s (b_0 + b_2 t^2 + b_4 t^4 + \ldots) = b_0 t(1+ P(t)), $$

where $P(t)$ is a polynomial.

You can now use variation of parameters as I did here to show that the solution can be written as follows:

$$x(t) = A y_1(t) + B y_2(t) \int \frac{dt}{u(t)},$$

where $A$ and $B$ are constants of integration and $u$ is an integrating factor given by:

$$u(t) = e^{\int \frac{2y_1'}{y_1} \, dt} = y_1^2.$$

Recall that you can perform the integral $\int \frac{dt}{y_1^2} = \frac{1}{b_0^2}\int \frac{dt}{t(1+P(t))}$ by taking into account the Taylor expansion for $(1+P)^{-1}$ which is:

$$\frac{1}{1+P} = 1-2P+3P^2-4P^3+\ldots$$

You should check uniform convergence of the series that give the solution.

I hope this may help you.

Good luck!

  • $\begingroup$ It would be much appreciated if you can try to solve my question, i will be around trying to solve it myself :) $\endgroup$
    – José D.
    Jan 28, 2014 at 11:21
  • $\begingroup$ I apologize for the continous editing. $\endgroup$
    – Dmoreno
    Jan 28, 2014 at 11:44
  • $\begingroup$ No problem, thanks for your help! $\endgroup$
    – José D.
    Jan 28, 2014 at 12:18

Let me just answer the first part here:

Your final expression can be re-written as $$ a_{k+2} = -\frac{k(k-1) - a(a+1)}{(k+2)(k+1)} a_k $$ Which in particular implies $$ a_{(a+1)+2} = 0 \implies a_{(a+1) + 2\ell} = 0 \qquad \forall \ell \in \{0,1,2,\dots\}$$

Note also that your recursion relation is such that $a_{k+2}$ is a function of $a_k$ only. This tells you that the coefficients for $k$ even and $k$ odd are independent. So if $a_0 = 0$, then all even coefficients must vanish; similarly for $a_1 = 0$ and all odd coefficients.

Now suppose $a+1$ is even. Then choosing $a_1 = 0$, we have that all odd coefficients must vanish. But by our first paragraph we also have that regardless of what $a_0$ is, all even coefficients for degree $k > a+1$ also must vanish. This tells you that there exists a solution that is a polynomial of degree $a+1$ if you choose $a_0 \neq 0$.

Similarly, suppose $a+1$ is odd. By choosing $a_0 = 0$ and $a_1 \neq 0$, you can conclude that there exists a polynomial solution of degree $a+1$.


I can be totally wrong here but let me give you my feeling : you have a second order differential equation without any boundary conditions. So, you can express all the coefficients as functions of $a_0$,$a_1$ and $a$. You have established the recurrence relaion.

  • $\begingroup$ Yes, but how can i get a general solution for that recurrence relation. I mean, how can i express $a_k$ as a function of $k, a_0,a_1$? $\endgroup$
    – José D.
    Jan 28, 2014 at 11:06

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