 Solution

Divide the Chebyshev differential equation by $1-x^2$. Its coefficients $-x/(1-x^2)$ and $m^2/(1-x^2)$ are analytic at zero, so $\boxed{x=0\text{ is an ordinary point}.}$ The finite singular points of a differential equation are $x=\pm 1$.

Insert $y={\sum }_{n\ge 0}{a}_n{x}^{n+\sigma }$, $a_0\ne 0$. The coefficient of the lowest power $x^{\sigma -2}$ is $\sigma (\sigma -1)a_0$, hence $\sigma =0$ or $1$. The next power gives $\sigma (\sigma +1)a_1=0$, and the remaining powers yieldFor $\sigma =0$, $a_1$ is free andChoosing $a_1=0$ selects the even power series, with $y(0)=a_0,y^{\prime }(0)=0$. For $\sigma =1$, the next-power equation forces $a_1=0$ andThis selects the odd power series, with $y(0)=0,y^{\prime }(0)=a_0$. Their different initial values prove linear independence. Although the Frobenius method produces two possible leading exponents, zero is an ordinary point, and a general $\sigma =0$ series already includes both initial constants.