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  "body": "\n## Proof\n\n1. The [[fundamental matrix|0MQ24RBG93]] $\\Phi$ is differentiable, so its entries are continuous on $I$. By the [[Leibniz formula|016PE4VZ8Q]], $\\det \\Phi(s)$ is a polynomial in these entries, so it is continuous in $s$.\n2. Each $\\Phi(s)$ is invertible, so [[$\\det \\Phi(s) \\ne 0$|03Q7RE2HXR]]. Let $\\Psi(s) = \\Phi(s)^{-1}$.\n3. For $n = 1$, $\\Psi = 1/\\Phi$ is continuous. For $n \\ge 2$, $\\Psi(s) = \\operatorname{adj} \\Phi(s) / \\det \\Phi(s)$ by [[the adjugate formula|0ENVM493VK]]. The entries of the adjugate are signed minors, which are polynomials in the entries of $\\Phi$. So $\\Psi$ is continuous on $I$.\n4. Let $g(s) = \\Psi(s) b(s)$, a continuous map $I \\to \\R^n$. Let $c(t) = \\Psi(t_0) x_0 + \\int_{t_0}^{t} g(s)\\, ds$, componentwise.\n5. By [[the antiderivative theorem|0NC2A3HWGT]], applied to each component, $c$ is differentiable on $I$ with $c' = g$.\n6. Let $x(t) = \\Phi(t) c(t)$. This is the formula of the statement.\n7. By the entrywise [[product rule|0V9A43H7WV]] and [[linearity of differentiation|0M0TKFRX2A]], $x' = \\Phi' c + \\Phi c' = A \\Phi c + \\Phi \\Psi b = A x + b$ on $I$.\n8. Also $x(t_0) = \\Phi(t_0) \\Psi(t_0) x_0 + 0 = x_0$. So $x$ solves the [[linear system|0A6QZ367GX]] $x' = A(t) x + b(t)$ on $I$ with $x(t_0) = x_0$.\n9. Uniqueness. Let $F(t, x) = A(t) x + b(t)$ on the open set $D = I \\times \\R^n$. It is continuous.\n10. Let $(t_1, x_1) \\in D$. Choose $[c, d] \\subseteq I$ with $c \u003c t_1 \u003c d$. By the [[extreme value theorem|0WS1A8E2P9]], the continuous function $t \\mapsto \\left(\\sum_{i, j} A_{ij}(t)^2\\right)^{1/2}$ has a maximum $K$ on $[c, d]$.\n11. For $t \\in [c, d]$ and $v \\in \\R^n$, the [[Cauchy–Schwarz inequality|0RFWW56Y22]] on each row gives $\\norm{A(t) v} \\le K \\norm{v}$. So $\\norm{F(t, x) - F(t, y)} \\le K \\norm{x - y}$ on $(c, d) \\times \\R^n$.\n12. Thus $F$ is locally [[Lipschitz|055VVSN00V]] in $x$. By [[uniqueness of solutions for Lipschitz vector fields|025PKW8AMR]], any solution on $I$ with value $x_0$ at $t_0$ equals $x$.\n\n## Idea\n\nTry $x = \\Phi c$ with a varying vector $c$. Since $\\Phi' = A \\Phi$, the equation reduces to $\\Phi c' = b$, so $c' = \\Phi^{-1} b$. Integrating $c'$ from $t_0$ gives the formula, and uniqueness holds because the right side is locally Lipschitz in $x$.\n",
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