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  "body": "\n## Statement\n\nLet $J$ be an interval, let $U$ be an open subset of $\\mathbb{R}^n$, let $f : J \\times U \\to \\mathbb{R}^n$ be continuous, and let $t_0 \\in J$ and $x_0 \\in U$.\nLet $\\gamma : J \\to U$ be a continuous curve.\nThen $\\gamma$ is differentiable on $J$ with $\\gamma'(t) = f(t, \\gamma(t))$ for each $t \\in J$ and $\\gamma(t_0) = x_0$ if and only if\n$$\\gamma(t) = x_0 + \\int_{t_0}^{t} f\\left(s, \\gamma(s)\\right) ds \\quad \\text{for each } t \\in J.$$\nAt an endpoint of $J$ that belongs to $J$, the derivative is one-sided.\n\n## Remarks\n\nIn other words, the solutions of the [[initial value problem|075VBY4K4X]] on $J$ are the fixed points of the [[Picard operator|00WAVB68F9]].\nProof: one direction is the fundamental theorem of calculus applied to $\\gamma'$. For the other, the integrand $s \\mapsto f(s, \\gamma(s))$ is continuous, so the integral is differentiable in $t$.\nThe integral form needs only continuity of $\\gamma$. This makes it the right setting for fixed point arguments.\n\n## Explanations\n\n### L1 human\n\nIntegrate both sides of $\\gamma' = f(t, \\gamma)$ from $t_0$ to $t$.\nThe initial condition and the differential equation become one equation with an integral.\n",
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