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  "body": "\n## Definition\n\nLet $k$ and $f$ be locally integrable functions on $[0, \\infty)$ whose [[Laplace transforms|0V49JEZW06]] $\\hat k(p) = \\int_0^\\infty e^{-pt} k(t) \\, dt$ and $\\hat f(p)$ converge absolutely for $\\operatorname{Re} p \u003e c$, for some real $c$.\nConsider the [[Volterra equation of convolution type|SNFYTC1TTA]]\n$$x(t) = f(t) + \\int_0^t k(t - s) \\, x(s) \\, ds, \\quad t \\ge 0.$$\nThe *Laplace transform method* for this equation consists of the following steps.\n\n1. Assume that the solution $x$ has a Laplace transform $\\hat x$, and apply the [[convolution theorem|0JNMJEDD93]] to get $\\hat x(p) = \\hat f(p) + \\hat k(p) \\, \\hat x(p)$.\n2. Solve: $\\hat x(p) = \\dfrac{\\hat f(p)}{1 - \\hat k(p)}$ for $\\operatorname{Re} p$ large enough that $\\hat k(p) \\ne 1$.\n3. Recover $x$ as the [[inverse Laplace transform|07WBTX1MHA]] of $\\hat x$, and check that it solves the equation.\n\nFor an equation of the first kind $\\int_0^t k(t - s) \\, x(s) \\, ds = f(t)$, step 2 becomes $\\hat x = \\hat f / \\hat k$.\n\n## Remarks\n\nExample: $x(t) = t - \\int_0^t (t - s) \\, x(s) \\, ds$, with $k(u) = -u$ and $f(t) = t$.\nHere $\\hat k(p) = -1/p^2$ and $\\hat f(p) = 1/p^2$, so $\\hat x(p) = \\frac{1/p^2}{1 + 1/p^2} = \\frac{1}{p^2 + 1}$ and $x = \\sin$.\nCheck: $\\int_0^t (t - s) \\sin s \\, ds = t - \\sin t$, so the right side is $\\sin t$.\nNon-example: the method does not apply directly to $x(t) = 1 + \\int_0^t ts \\, x(s) \\, ds$, since the kernel $ts$ is not a function of $t - s$, and the transform of the integral term is not a product.\nThe step 1 assumption is harmless when $k$ and $f$ are as above: the unique locally integrable solution then has a Laplace transform for $\\operatorname{Re} p$ large.\nThe function $r$ with $\\hat r = \\hat k / (1 - \\hat k)$ is the resolvent kernel, and $x = f + r * f$.\n",
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