From a problem I have I derived the following equation
$$ \left\{ \begin{array}{l} \left(2k + 1\right)xP_k(x) + (1-x^2)\frac{d}{dx}P_k(x) = P_{k+1}(x) \\ P_1(x)=1 \end{array} \right. . $$
The equation is derived by my guess that for all $k \geq 1$ $$ \frac{d^k}{dx^k} \arcsin(x) = \frac{1}{\left(1-x^2\right)^{k+\frac{1}{2}}} P_k(x) $$ Where $P_k(x)$ is a polynomial of degree $k-1$. (I suppose that you can prove it by induction).
Is there a way you can suggest me to solve that equation? The purpose would be basically find all the coefficients of $P_k$ for given $k$.
Update : Attempt I made
I think I've derived a recurrence relationship for $P_k(x)$ Let's consider the relationship
$$ \left(2k - 1\right)xP_{k-1}(x) + (1-x^2)\frac{d}{dx}P_{k-1}(x) = P_{k}(x) $$
Let's express $$ P_k(x)= \sum_{j=-\infty}^{+\infty}a_{j,k}x^j, \;\; a_{j,k}=0 \;\; \forall j < 0, j > k - 1 $$ Substituing in recurrence we get $$ \sum_{j=-\infty}^{+\infty} \left[ \left(2k -j\right)a_{j-1,k-1} + \left(j+1 \right)a_{j+1,k-1}\right]x^j = \sum_{j=-\infty}^{+\infty}a_{j,k}x^j $$ To be equal the two polynomial it must be verified that $$ \left(2k -j\right)a_{j-1,k-1} + \left(j+1 \right)a_{j+1,k-1} = a_{j,k} $$
What puzzles me in the relantionship above is that the $j$-th coefficient of $P_k$ depends from the coefficient $j+1$-th of $P_{k-1}$ do you think there's some mistakes in my derivation?
I used series instead of summation because it makes easier to handle boundary conditions.