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  "body": "\n## Summary\n\nTo solve an [[initial value problem|075VBY4K4X]] for a [[linear equation with constant coefficients|0A6PGTTP2A]], one applies the [[Laplace transform|05XT5AVJBK]] to both sides of the equation.\nThe [[rule for derivatives|0K090GVHVN]] turns the equation and the initial values into one algebraic equation for the transform $Y(s)$ of the unknown $y(t)$.\nOne solves this equation for $Y(s)$ and then finds $y = \\mathcal{L}^{-1}\\{Y\\}$ with the [[inverse Laplace transform|0JRW4X1MDC]].\n\n## Remarks\n\nExample: for $y'' + y = 0$, $y(0) = 0$, $y'(0) = 1$, the transform gives $s^2 Y(s) - 1 + Y(s) = 0$. So $Y(s) = 1/(s^2 + 1)$ and $y(t) = \\sin t$.\nFor $a_n y^{(n)} + \\cdots + a_0 y = g(t)$ with zero initial values, $Y(s) = G(s) / P(s)$, where $P(s) = a_n s^n + \\cdots + a_0$ and $G$ is the transform of $g$. The [[convolution theorem|0JNMJEDD93]] gives $y = h * g$ with $h = \\mathcal{L}^{-1}\\{1/P\\}$.\nThe method handles discontinuous forcing, such as a [[step function|09C4JP7Y0M]], and impulses, such as a [[Dirac delta|0KPHENZ5S7]].\n\n## Explanations\n\n### L1 human\n\nThe Laplace transform turns a differential equation into an algebra problem.\nWe solve the algebra problem, which is often easy, and then transform back to get the solution.\nThe initial values are built in from the start, so there are no constants to find at the end.\n",
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