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  "body": "\n## Statement\n\nLet $n \\ge 1$, let $U$ be a subset of $\\mathbb{R} \\times \\mathbb{R}^n$, and let $f : U \\to \\mathbb{R}$ be a function.\nConsider the equation $y^{(n)} = f\\left(x, y, y', \\dots, y^{(n-1)}\\right)$ and the first-order system for $z = (z_1, \\dots, z_n)$:\n$$z_1' = z_2, \\quad z_2' = z_3, \\quad \\dots, \\quad z_{n-1}' = z_n, \\quad z_n' = f(x, z_1, \\dots, z_n).$$\nLet $I$ be an open interval.\nIf $y$ is a [[solution|04SW7WECS8]] of the equation on $I$, then $z = \\left(y, y', \\dots, y^{(n-1)}\\right)$ is a solution of the system on $I$.\nConversely, let $z$ be a solution of the system on $I$. Then $y = z_1$ is a solution of the equation on $I$, and $z_k = y^{(k-1)}$ for $k = 1, \\dots, n$.\n\n## Remarks\n\nThe initial conditions $y^{(k)}(x_0) = c_k$ for $k = 0, \\dots, n-1$ become the one condition $z(x_0) = (c_0, \\dots, c_{n-1})$. So results on first-order [[initial value problems|075VBY4K4X]] apply to equations of each order.\nFor $y'' = -y$, the system is $z_1' = z_2$, $z_2' = -z_1$.\nThe same method changes a system of equations of order $n$ into a larger first-order system.\n\n## Explanations\n\n### L1 human\n\nThe trick is to give new names to the derivatives of $y$.\nEach new unknown is the derivative of the one before it, and the last one is given by the equation.\nSo we only need a theory of first-order systems.\n",
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