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  "body": "\n## Statement\n\nLet $I$ be an open interval, let $a_0, \\dots, a_{n-1} : I \\to \\mathbb{R}$ be continuous, and let $y_1, \\dots, y_n$ be solutions on $I$ of the [[homogeneous linear equation|0EQZSAB9EC]] $y^{(n)} + a_{n-1}(x) y^{(n-1)} + \\dots + a_0(x) y = 0$.\nLet $W = W(y_1, \\dots, y_n)$ be their [[Wronskian|0N5NF1RGBP]].\nThen these three conditions are equivalent.\n(a) The functions $y_1, \\dots, y_n$ are linearly independent on $I$.\n(b) There is a point $x_0 \\in I$ with $W(x_0) \\ne 0$.\n(c) For each $x \\in I$, $W(x) \\ne 0$.\n\n## Remarks\n\nSo the Wronskian of $n$ solutions is either zero at each point of $I$ or zero at no point of $I$. [[Abel's identity|0VGWKX2JHB]] gives the same result for $n = 2$.\nFor arbitrary functions, $W(x_0) \\ne 0$ at one point still implies linear independence. But the converse fails: $x^2$ and $x |x|$ are linearly independent on $\\mathbb{R}$, and their Wronskian is $0$ at each point.\nExample: $e^{x}$ and $e^{-x}$ solve $y'' - y = 0$, and their Wronskian is $-2$, so they are independent.\n\n## Explanations\n\n### L1 human\n\nTo check that $n$ solutions are independent, compute one determinant at one point.\nIf it is not zero, the solutions are independent and form a basis of all solutions.\nFor solutions of a linear equation, the value at one point already decides the question.\n",
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