# Uniqueness of a second order linear ode

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.

• What is $n$ here? If e.g. $n=1$, then any constant is a solution. Commented Feb 1, 2023 at 21:22

$$w \! \left(t \right) = c_{1} \mathit{HeunC} \left(2 n -1, n -1, n -2, -n^{2}+\frac{1}{2}, \frac{n^{2}}{2}+\frac{1}{2}, t\right)+c_{2} t^{-n +1} \mathit{HeunC} \left(2 n -1, -n +1, n -2, -n^{2}+\frac{1}{2}, \frac{n^{2}}{2}+\frac{1}{2}, t\right)$$ 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.
• No need to take real part. If $n$ is real and $t \in (-1,1)$, $\mathit{HeunC} \left(2 n -1, n -1, n -2, -n^{2}+\frac{1}{2}, \frac{n^{2}}{2}+\frac{1}{2}, t\right)$ is real. Commented Feb 1, 2023 at 22:28
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 $$n\geq 2$$.