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  "body": "\n## Proof\n\n1. By the definition of a [[solution|04SW7WECS8]] of the [[equation|0P97V20RW7]], $y : I \\to \\R$ solves $y^{(n)} = f\\left(x, y, \\dots, y^{(n-1)}\\right)$ on $I$ when three things hold. The function $y$ is $n$ times differentiable on $I$. For each $x \\in I$, the point $\\left(x, y(x), \\dots, y^{(n-1)}(x)\\right)$ lies in $U$. For each $x \\in I$, $y^{(n)}(x) = f\\left(x, y(x), \\dots, y^{(n-1)}(x)\\right)$.\n2. In the same way, $z = (z_1, \\dots, z_n) : I \\to \\R^n$ solves the system on $I$ when three things hold. The function $z$ is differentiable on $I$. For each $x \\in I$, $(x, z(x)) \\in U$, and the $n$ equations of the system hold at $x$.\n3. Let $y$ solve the equation on $I$, and set $z_k = y^{(k-1)}$ for $k = 1, \\dots, n$. Since $k - 1 \\le n - 1$, each $z_k$ is differentiable on $I$, with $z_k' = y^{(k)}$.\n4. For $k = 1, \\dots, n - 1$, this gives $z_k' = y^{(k)} = z_{k+1}$.\n5. For $k = n$, step 1 gives $z_n' = y^{(n)} = f\\left(x, y, \\dots, y^{(n-1)}\\right) = f(x, z_1, \\dots, z_n)$ on $I$.\n6. Also $(x, z(x)) = \\left(x, y(x), \\dots, y^{(n-1)}(x)\\right) \\in U$ for each $x \\in I$. By step 2, $z$ solves the system on $I$.\n7. Conversely, let $z$ solve the system on $I$, and set $y = z_1$. We show by induction on $k$ that $y$ is $k - 1$ times differentiable with $y^{(k-1)} = z_k$, for $k = 1, \\dots, n$.\n8. For $k = 1$ this is the definition $y = z_1$.\n9. Let $1 \\le k \u003c n$ and assume $y^{(k-1)} = z_k$. The function $z_k$ is differentiable with $z_k' = z_{k+1}$. So $y^{(k-1)}$ is differentiable and $y^{(k)} = z_{k+1}$. This completes the induction.\n10. In particular $y^{(n-1)} = z_n$, and $z_n$ is differentiable. So $y$ is $n$ times differentiable on $I$, and $y^{(n)} = z_n'$.\n11. By the last equation of the system, $y^{(n)} = z_n' = f(x, z_1, \\dots, z_n) = f\\left(x, y, \\dots, y^{(n-1)}\\right)$ on $I$.\n12. By steps 7 to 9, $\\left(x, y(x), \\dots, y^{(n-1)}(x)\\right) = (x, z(x)) \\in U$ for each $x \\in I$. So $y$ solves the equation on $I$, and $z_k = y^{(k-1)}$ for $k = 1, \\dots, n$.\n\n## Idea\n\nThe new unknowns $z_k$ are names for the derivatives $y^{(k-1)}$. The first $n - 1$ equations of the system say exactly that each name is the derivative of the previous one. The last equation is the original equation, written in the new names.\n",
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