Uniqueness of a second order linear ode

Question

I am currently confronted with the following equation $$ 0=w''(t)(t^2-t)+w'(t)((2n-1)t^2-n)+w(t)(n-1)^2t $$ for $t\in(-1,1)$. So $w:(-1,1)\rightarrow\mathbb{R}$. The following assumption is also in force, $w(t)\in C^{2}(-1,1)$

I would like to prove that the solution to the above problem is unique (w=0). I have found some papers which prove uniqueness of such a class of equations with coefficients that contain singularities but none that are applicable in my case. Does anyone have an idea?

Edit: I apologise for not having added the following. n is an integer greater than or equal to 2.

Answer 1

It is not unique. Maple solves your differential equation in terms of HeunC functions:

$$w \! \left(t \right) = c_{1} \mathit{HeunC} \left(2 n -1, n -1, n -2, -n^{2}+\frac{1}{2}, \frac{n^{2}}{2}+\frac{1}{2}, t\right)+c_{2} t^{-n +1} \mathit{HeunC} \left(2 n -1, -n +1, n -2, -n^{2}+\frac{1}{2}, \frac{n^{2}}{2}+\frac{1}{2}, t\right)$$ The first basic solution $$ \mathit{HeunC} \left(2 n -1, n -1, n -2, -n^{2}+\frac{1}{2}, \frac{n^{2}}{2}+\frac{1}{2}, t\right)$$ is analytic in the open unit disk.

Answer 2

The set of $C^2$ solutions is one-dimensional, that is "unique up to a constant factor". The exponents at $0$ are $0$ and $1-n$, which gives a basis $w_1,w_2$ of solutions where $w_1$ is holomorphic, and $w_2$ blows up since $n\geq 2$.