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  "body": "\n## Statement\n\nLet $I$ be an open interval, let $p, q : I \\to \\mathbb{R}$ be continuous, let $x_0 \\in I$ and let $y_0 \\in \\mathbb{R}$.\nLet $P(x) = \\int_{x_0}^{x} p(s)\\, ds$ for $x \\in I$.\nThen the [[initial value problem|075VBY4K4X]] $y' + p(x)\\, y = q(x)$, $y(x_0) = y_0$ has exactly one solution on $I$, and it is\n$$y(x) = e^{-P(x)} \\left( y_0 + \\int_{x_0}^{x} e^{P(s)}\\, q(s)\\, ds \\right), \\qquad x \\in I.$$\n\n## Remarks\n\nProof idea: $\\mu = e^{P}$ is an [[integrating factor|01THH3Y3YP]], so a solution satisfies $(\\mu y)' = \\mu q$. Integrate from $x_0$ to $x$ and use $\\mu(x_0) = 1$.\nThe solution exists on all of $I$. A linear equation has no solution that blows up inside the interval of continuity of $p$ and $q$.\nExample: for $y' + y = 1$, $y(0) = 0$, the formula gives $y(x) = 1 - e^{-x}$.\n\n## Explanations\n\n### L1 human\n\nThe formula has two parts.\nThe term $y_0 e^{-P(x)}$ is the solution of the equation without the right side $q$.\nThe integral term adds the effect of $q$, collected over the time from $x_0$ to $x$.\n",
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