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  "body": "\n## Definition\n\nLet $I = [t_0, t_0 + T]$ with $T \u003e 0$, let $n \\ge 1$, and let $f \\colon I \\times \\R^n \\to \\R^n$ and $K \\colon I \\times I \\times \\R^n \\to \\R^n$ be continuous.\nA (Volterra type) *integro-differential equation* is an equation\n$$x'(t) = f(t, x(t)) + \\int_{t_0}^{t} K(t, s, x(s)) \\, ds, \\quad t \\in I,$$\nfor an unknown continuously differentiable $x \\colon I \\to \\R^n$.\nMore generally, an integro-differential equation is any equation with the unknown both under a derivative and under an [[integral|0W29TBTY12]].\nWhen the integral is over a fixed interval, the equation is of Fredholm type.\n\n## Remarks\n\nExample: $x'(t) = -\\int_0^t x(s) \\, ds$, $x(0) = 1$.\nDifferentiating gives $x'' = -x$ with $x(0) = 1$ and $x'(0) = 0$, so $x = \\cos t$.\nCheck: $-\\int_0^t \\cos s \\, ds = -\\sin t = x'(t)$.\n\nNon-example: $x'(t) = -x(t)$ contains no integral of the unknown, so it is an [[ordinary differential equation|0NT64JV96A]].\nAlso $x(t) = 1 + \\int_0^t x(s) \\, ds$ contains no derivative; it is a [[Volterra integral equation|DND8CBEVXT]].\n\nIntegrating the Volterra form from $t_0$ to $t$ turns the problem with $x(t_0) = x_0$ into a Volterra integral equation for $x$.\nFor a convolution kernel $K(t, s, x) = k(t - s) x$, the Laplace transform turns the linear equation into an algebraic one.\n\n## Explanations\n\n### L1 human\n\nThe rate of change is given partly by the present value and partly by a weighted sum of all past values.\n",
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