Question: The Jacobi differential equation in terms of the Jacobi polynomial $P_{n}^{(\alpha,0)}(x)$ is given by: \begin{equation} (1 - x^2) P_{n}^{(\alpha,0)''}(x) + (-\alpha - (\alpha + 2) x) P_{n}^{(\alpha,0)'}(x) + n(n + \alpha + 1). P_{n}^{(\alpha,0)}(x) = 0 \;\;\; (*) \end{equation} Prove that $T_{n}(x)=x^{k+\frac{1}{2}} P_{n}^{(\alpha,0)}(1-2x^2) $ satisfies in the following ODE \begin{equation} (1 - x^2) T_{n}''(x) -2x T_{n}'(x) + \frac{\frac{1}{4}-k^2}{x^{2}} T_{n}(x)= -\chi T_{n}(x). \;\;\; (**) \end{equation}
I understand that $$ T_{n}'(x) = (k+\frac{1}{2})x^{k-\frac{1}{2}} P_{n}^{(\alpha,0)}(1-2x^2) - 4x^{k+\frac{3}{2}} P_{n}^{(\alpha,0)'}(1-2x^2) $$ and $T_{n}''$, however when I substitute $T_{n},\; T_{n}',\; T_{n}''$ in (**) I don't get anything close to (*). I have also thought about the change of variable $t=1-2x^{2}$, but no result. I was wondering if someone could point out my mistake about this.
In order to simplify notation, let's write $T(x)$ and $P(x)$ instead of $T_n(x)$ and $P_n^{(\alpha,0)}(x)$, respectively. Let's also define $q=k+\frac{1}{2}$ and $t=1-2x^2$. In this notation, we have $T(x)=x^qP(t)$, hence \begin{align} T'(x)&=x^qP'(t)\frac{dt}{dx}+qx^{q-1}P(t) \\ &=-4x^{q+1}P'(t)+qx^{q-1}P(t), \tag{1} \\ T''(x)&=-4x^{q+1}P''(t)\frac{dt}{dx}-4(q+1)x^qP'(t)+qx^{q-1}P'(t)\frac{dt}{dx}+ q(q-1)x^{q-2}P(t) \\ &=16x^{q+2}P''(t)-4(2q+1)x^qP'(t)+q(q-1)x^{q-2}P(t). \tag{2} \end{align} Let's now plug $(1)$ and $(2)$ into the LHS of your equation (**): \begin{align} \text{LHS } (**)&=(1-x^2)T''(x)-2xT'(x)+\frac{\frac{1}{4}-k^2}{x^2}T(x) \\ &=(1-x^2)16x^{q+2}P''(t)+\left[-(1-x^2)4(2q+1)x^q+8x^{q+2}\right]P'(t) \\ &\quad+\left[(1-x^2)q(q-1)x^{q-2}-2qx^q-q(q-1)x^{q-2}\right]P(t), \tag{3} \end{align} where we used the identity $k^2-\frac{1}{4}=q(q-1)$ in the last term of $(3)$. To further simplify $(3)$, we shall use the identities $x^2=\frac{1-t}{2}$ and $1-x^2=\frac{1+t}{2}$: \begin{align} \text{LHS } (**)&=4(1-t^2)x^qP''(t)+[-4q+2-(4q+6)t]x^qP'(t)-q(q+1)x^qP(t) \\ &=4x^q\left[(1-t^2)P''(t)+(-k-(k+2))P'(t)\right]-q(q+1)x^qP(t). \tag{4} \end{align} Now, if $k=\alpha$, we can use your equation $(*)$ to replace the terms in square brackets with $-n(n+\alpha+1)P(t)$, so that $$ \text{LHS } (**)=-[4n(n+\alpha+1)+q(q+1)]x^qP(t)=-\chi T(x), \tag{5} $$ where, using $q=k+\frac{1}{2}=\alpha+\frac{1}{2}$, we can rewrite $\chi$ as $$ \chi=4n(n+\alpha+1)+\left(\alpha+\frac{1}{2}\right)\left(\alpha+\frac{3}{2}\right). \tag{6} $$ In conclusion, $T_n(x)=x^{k+\frac{1}{2}}P_n^{(\alpha,0)}(1-2x^2)$ satisfies your equation $(**)$ if $k=\alpha$ and $\chi$ is given by $(6)$.