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?