Find the general solution of the Chebyshev equation…

Find the general solution of the Chebyshev equation $(1-x^2)y'' - xy' + n^2y = 0, \qquad n\in\mathbb{R},\ \ |x|\lt 1$.

Step 1: The trigonometric substitution $x=\cos\theta$

Since $|x|\lt 1$, set $x=\cos\theta$ with $\theta\in(0,\pi)$, and write $\dot y = dy/d\theta$. From $dx/d\theta=-\sin\theta$, \[y' = \frac{dy}{dx} = \frac{\dot y}{dx/d\theta} = -\frac{\dot y}{\sin\theta}.\]

Differentiating again with respect to $x$ (chain rule through $\theta$) gives

\[ y'' = \frac{d}{d\theta}\!\left(-\frac{\dot y}{\sin\theta}\right)\cdot\frac{d\theta}{dx} = -\frac{\ddot y\sin\theta - \dot y\cos\theta}{\sin^2\theta}\cdot\left(-\frac{1}{\sin\theta}\right) = \frac{\ddot y\sin\theta - \dot y\cos\theta}{\sin^3\theta}. \]

Step 2: The transformed equation collapses to $\ddot y + n^2y = 0$

Using $1-x^2=\sin^2\theta$ and $x=\cos\theta$, \[(1-x^2)y'' = \sin^2\theta\cdot\frac{\ddot y\sin\theta-\dot y\cos\theta}{\sin^3\theta} = \ddot y - \dot y\,\frac{\cos\theta}{\sin\theta}, \qquad -xy' = -\cos\theta\left(-\frac{\dot y}{\sin\theta}\right) = \dot y\,\frac{\cos\theta}{\sin\theta}.\]

Adding these two pieces, the $\dot y\cos\theta/\sin\theta$ terms cancel exactly, leaving the strikingly simple

\[ \ddot y + n^2y = 0. \]

Step 3: Solving in $\theta$

Step 4: Back-substitution and the Chebyshev polynomials

Returning to $\theta=\arccos x$, the general solution for $n\neq0$ is

(for $n=0$: $y=C_1+C_2\arccos x$). The first solution, $T_n(x)=\cos(n\arccos x)$, is the Chebyshev polynomial of the first kind whenever $n$ is a nonnegative integer — e.g. $T_0=1$, $T_1=x$, $T_2=2x^2-1$, $T_3=4x^3-3x$ — obtained from the multiple-angle expansion of $\cos(n\theta)$ in powers of $\cos\theta$. The second solution $\sin(n\arccos x)$ is likewise polynomial in $x$ after dividing by $\sin\theta=\sqrt{1-x^2}$; that quotient, $U_{n-1}(x)=\sin(n\theta)/\sin\theta$, is the Chebyshev polynomial of the second kind, but as it stands $\sin(n\arccos x)$ is already a perfectly good (non-polynomial, but smooth on $(-1,1)$) second solution.

Step 5: Verification

For $y_1=\cos(n\theta)$: $\dot y_1=-n\sin(n\theta)$, $\ddot y_1=-n^2\cos(n\theta)$, so $\ddot y_1+n^2y_1=-n^2\cos(n\theta)+n^2\cos(n\theta)=0.\ \checkmark$ Likewise for $y_2=\sin(n\theta)$, $\ddot y_2+n^2y_2=0.\ \checkmark$

Their Wronskian in $\theta$ is

\[ W(y_1,y_2) = y_1\dot y_2 - \dot y_1y_2 = n\cos^2(n\theta) + n\sin^2(n\theta) = n, \]

which is nonzero exactly when $n\neq0$, confirming that $\cos(n\theta)$ and $\sin(n\theta)$ are independent solutions in that case (and explaining why the $n=0$ case needs the separate fundamental pair $\{1,\theta\}$).