ordinary differential equations - Problem about ODE and power series

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?

Answer

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 $a+1$.