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  "body": "\n## Example\n\nLet $\\lambda \\in \\C$. The *Hermite equation* is\n$$y'' - 2 z\\, y' + 2 \\lambda\\, y = 0.$$\nIts coefficients $-2z$ and $2 \\lambda$ are polynomials, so every point of $\\C$ is an [[ordinary point|QK2J038VB4]].\nA [[power series solution|08RT9GDDPQ]] $y = \\sum_{k \\ge 0} a_k z^k$ gives\n$$(k + 2)(k + 1)\\, a_{k+2} - 2 k\\, a_k + 2 \\lambda\\, a_k = 0, \\quad \\text{so} \\quad a_{k+2} = \\frac{2 (k - \\lambda)}{(k + 2)(k + 1)}\\, a_k.$$\nThe values $a_0$ and $a_1$ are free, and every solution is an entire function.\nIf $\\lambda = n$ is a nonnegative integer, the recurrence stops at $k = n$ for the series of the same parity as $n$.\nThis gives a polynomial solution of degree $n$, a multiple of the [[Hermite polynomial|MGACY98E9H]] $H_n$.\n\n## Remarks\n\nFor $n = 2$, $H_2(z) = 4 z^2 - 2$; then $y'' = 8$ and $y' = 8 z$, and $8 - 2 z \\cdot 8 z + 4 (4 z^2 - 2) = 0$.\nThe substitution $y = e^{z^2 / 2} u$ links the equation to the quantum harmonic oscillator $u'' + (2 \\lambda + 1 - z^2) u = 0$.\nThe point $\\infty$ is an irregular singular point.\n\n## Explanations\n\n### L1 human\n\nThe Hermite equation has no bad points in the finite plane, so its solutions are power series that converge everywhere.\nFor whole numbers $\\lambda = n$ one of the series stops and becomes a polynomial.\nThese Hermite polynomials describe the energy states of a quantum harmonic oscillator.\n",
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