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  "body": "\n## Proof\n\nWe reduce to a first-order linear system and use existence and uniqueness for it.\n\n1. The set $S$ consists of the solutions of the [[homogeneous linear equation|0EQZSAB9EC]] on $I$. The zero function lies in $S$. By the [[superposition principle|0F2VBPXNP9]], $S$ is closed under linear combinations. So $S$ is a subspace of the real vector space of functions $I \\to \\R$.\n2. Fix $x_0 \\in I$ and let $\\Phi(y) = \\left(y(x_0), y'(x_0), \\dots, y^{(n-1)}(x_0)\\right)$ for $y \\in S$. By [[linearity of differentiation|0M0TKFRX2A]], as in the superposition principle, $\\Phi : S \\to \\R^n$ is linear.\n3. For $x \\in I$, let $A(x)$ be the $n \\times n$ matrix with entries $A_{k, k+1}(x) = 1$ for $k \u003c n$, last row $\\left(-a_0(x), \\dots, -a_{n-1}(x)\\right)$, and all other entries $0$. Each entry of $A$ is continuous on $I$.\n4. Apply the [[reduction to a first-order system|0TH9CPW3VS]] with $U = I \\times \\R^n$ and $f(x, z) = -a_0(x) z_1 - \\dots - a_{n-1}(x) z_n$. The system becomes the [[linear system|0A6QZ367GX]] $z' = A(x) z$.\n5. By that reduction, $y \\in S$ implies that $z = \\left(y, \\dots, y^{(n-1)}\\right)$ solves $z' = A(x) z$ on $I$. Conversely, each solution $z$ on $I$ has $z_1 \\in S$ and $z = \\left(z_1, \\dots, z_1^{(n-1)}\\right)$.\n6. Let $[c, d] \\subseteq I$ be a closed bounded interval. The function $x \\mapsto \\left(\\sum_{i, j} A_{ij}(x)^2\\right)^{1/2}$ is continuous. By the [[extreme value theorem|0WS1A8E2P9]], it has a maximum $K$ on $[c, d]$.\n7. Let $x \\in [c, d]$ and $v \\in \\R^n$. Entry $i$ of $A(x) v$ is the dot product of row $i$ of $A(x)$ with $v$. The [[Cauchy–Schwarz inequality|0RFWW56Y22]] bounds its square by $\\left(\\sum_j A_{ij}(x)^2\\right) \\norm{v}^2$.\n8. Summing over $i$ gives $\\norm{A(x) v} \\le K \\norm{v}$. So $\\norm{A(x) z - A(x) w} \\le K \\norm{z - w}$ for $x \\in [c, d]$ and $z, w \\in \\R^n$, a [[Lipschitz condition|055VVSN00V]] in $z$.\n9. Injectivity. Let $y \\in S$ with $\\Phi(y) = 0$, and let $z = \\left(y, \\dots, y^{(n-1)}\\right)$. Then $z$ and the zero curve both solve $z' = A(x) z$ on $I$, and both vanish at $x_0$.\n10. The field $(x, z) \\mapsto A(x) z$ is continuous on the open set $D = I \\times \\R^n$. Each point $(x_1, z_1) \\in D$ has the neighbourhood $(c, d) \\times \\R^n$, where $[c, d] \\subseteq I$ and $c \u003c x_1 \u003c d$. By step 8, the field is Lipschitz in $z$ there.\n11. By [[uniqueness of solutions for Lipschitz vector fields|025PKW8AMR]], $z = 0$ on $I$. So $y = z_1 = 0$. The kernel of $\\Phi$ is trivial, so [[$\\Phi$ is injective|0BJBYCGMWE]].\n12. Surjectivity. Let $c \\in \\R^n$. Write $I = (\\alpha, \\beta)$ with $-\\infty \\le \\alpha \u003c \\beta \\le \\infty$. Choose numbers $c_m$ decreasing to $\\alpha$ and $d_m$ increasing to $\\beta$, with $c_1 \u003c x_0 \u003c d_1$.\n13. Let $J_m = [c_m, d_m] \\subseteq I$. By step 8 with $[c, d] = J_m$, the field $A(x) z$ is continuous on $J_m \\times \\R^n$ and Lipschitz in $z$ on all of it.\n14. By [[global existence for globally Lipschitz vector fields|0BFD3A6CRH]], there is exactly one solution $z_m : J_m \\to \\R^n$ of $z' = A(x) z$ with $z_m(x_0) = c$.\n15. Let $m \\le k$. The restriction of $z_k$ to $J_m$ is a solution on $J_m$ with value $c$ at $x_0$. By the uniqueness in step 14, it equals $z_m$.\n16. For $x \\in I$, let $z(x) = z_m(x)$ for any $m$ with $x \\in J_m$. Such $m$ exists since $c_m \\to \\alpha$ and $d_m \\to \\beta$. By step 15, $z(x)$ does not depend on $m$.\n17. Each $x \\in I$ lies in some open interval $(c_m, d_m)$, and $z = z_m$ there. So $z$ is differentiable at $x$ with $z'(x) = A(x) z(x)$. Also $z(x_0) = c$.\n18. By step 5, $y = z_1$ lies in $S$, and $\\Phi(y) = z(x_0) = c$. So $\\Phi$ is surjective.\n19. Thus $\\Phi$ is a linear bijection, and its inverse $\\Phi^{-1}$ is linear. Let $e_1, \\dots, e_n$ be the standard basis of $\\R^n$, and let $b_k = \\Phi^{-1}(e_k)$.\n20. If $\\sum_k \\lambda_k b_k = 0$, applying $\\Phi$ gives $\\sum_k \\lambda_k e_k = 0$, so each $\\lambda_k = 0$. Each $y \\in S$ equals $\\Phi^{-1}(\\Phi(y)) = \\sum_k \\Phi(y)_k\\, b_k$.\n21. So $(b_1, \\dots, b_n)$ is a [[basis|0KCQ9Y3QKA]] of $S$. Hence $S$ is a real vector space of [[dimension|05G2G0P224]] $n$, and $\\Phi$ is a linear bijection.\n\n## Idea\n\nA solution is fixed by its $n$ initial values $y(x_0), \\dots, y^{(n-1)}(x_0)$, and every choice of these values occurs. Uniqueness gives injectivity and existence gives surjectivity. Existence on all of $I$ holds because the equation is linear: on each closed subinterval, the field is globally Lipschitz.\n",
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