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Because zero is an ordinary point, substitute $y={\sum }_{j\ge 0}{a}_j{x}^j$. Matching powers givesEven and odd coefficients evolve independently. Taking $(a_0,a_1)=(1,0)$ and $(0,1)$ constructs solutionsTheir initial values and derivatives give $W(y_{\mathrm{e}},y_{\mathrm{o}})(0)=1$, so they are linearly independent. These power series of the Chebyshev differential equation span the local solution space. A useful check is $y_{\mathrm{e}}=\cos (n\arcsin x)$ and $y_{\mathrm{o}}=\sin (n\arcsin x)/n$, with $y_{\mathrm{o}}=\arcsin x$ at $n=0$.

The origin is an ordinary point. Substitute $y={\sum }_{n\ge 0}{a}_n{x}^n$ into the equation and collect the coefficient of $x^n$:Hence the recurrence relation isThere are two free initial coefficients. The independent even and odd power series beginThe general solution is $\boxed{y=a_0{y}_{\mathrm{e}}+a_1{y}_{\mathrm{o}}}$. As a useful independent check, the substitution $x=\sin t$ reduces the equation to $d^{2}y/d{t}^2+k^{2}y=0$. ThusFor a nonterminating parity series, $|a_{n+2}/a_n|\to 1$, so the ratio test in $x^2$ gives $\boxed{R=1}$. The regular singular points at $\pm 1$ explain the natural obstruction.

There is an important exception to an unconditional radius-one answer. If $k^2=m^2$ for an integer $m\ge 0$, the series of parity $m$ terminates at degree $m$ and is a multiple of a Chebyshev polynomial. Its radius of convergence is $\boxed{R=\infty }$; the other independent parity solution still has radius one. For $m=0$ the even solution is constant. Any nonzero combination containing the nonterminating solution has radius one. These are the full power series of the Chebyshev differential equation, including the terminating cases.