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  "body": "\n## Proof\n\nWe use the integrating factor $\\mu = e^{P}$.\n\n1. The function $p$ is continuous on $I$. So $P(x) = \\int_{x_0}^{x} p(s)\\, ds$ is differentiable on $I$ with $P' = p$, by [[the integral of a continuous function is an antiderivative|0NC2A3HWGT]]. Also $P(x_0) = 0$.\n2. Let $\\mu(x) = e^{P(x)}$. By the [[chain rule|0RXX2PT1MQ]] and [[the derivative of the exponential|0EX9GFJ0B3]], $\\mu' = p\\, \\mu$ on $I$.\n3. The [[exponential is positive|0P9BNBW2W3]], so $\\mu(x) \u003e 0$ for each $x \\in I$. Also $\\mu(x_0) = e^0 = 1$. So $\\mu$ is an [[integrating factor|01THH3Y3YP]] of the [[linear first-order equation|0Z3YKRQZ48]].\n4. Let $y : I \\to \\R$ be differentiable. By the [[product rule|0V9A43H7WV]] and step 2, $(\\mu y)' = \\mu y' + p \\mu y = \\mu\\, (y' + p y)$.\n5. The function $\\mu q$ is continuous on $I$. So $Q(x) = \\int_{x_0}^{x} \\mu(s)\\, q(s)\\, ds$ is differentiable on $I$ with $Q' = \\mu q$, again by the antiderivative theorem. Also $Q(x_0) = 0$.\n6. Uniqueness. Let $y$ be a solution of the [[initial value problem|075VBY4K4X]] on $I$. By step 4, $(\\mu y)' = \\mu q = Q'$.\n7. So $G = \\mu y - Q$ has $G' = 0$ on $I$, by [[linearity of differentiation|0M0TKFRX2A]]. Hence [[$G$ is constant|003QDZ4T16]] on $I$.\n8. Its value at $x_0$ is $\\mu(x_0) y(x_0) - Q(x_0) = y_0$. So $\\mu(x) y(x) = y_0 + Q(x)$ for each $x \\in I$.\n9. Divide by $\\mu(x) \u003e 0$ and use $1/\\mu(x) = e^{-P(x)}$. This gives the formula $y(x) = e^{-P(x)}\\left(y_0 + Q(x)\\right)$ of the statement. So there is at most one solution.\n10. Existence. Define $y(x) = e^{-P(x)}\\left(y_0 + Q(x)\\right)$ for $x \\in I$. It is differentiable on $I$ by steps 1 and 5, the chain rule and the product rule.\n11. Here $\\mu y = y_0 + Q$, so $(\\mu y)' = Q' = \\mu q$. By step 4, $\\mu\\, (y' + p y) = \\mu q$ on $I$.\n12. Dividing by $\\mu \u003e 0$ gives $y' + p y = q$ on $I$. Also $y(x_0) = e^{0}(y_0 + 0) = y_0$.\n13. So $y$ is a solution on $I$, and by step 9 it is the only one.\n\n## Idea\n\nMultiplying by $\\mu = e^{P}$ turns the left side $y' + p y$ into the exact derivative $(\\mu y)'/\\mu$. Then the equation says that $\\mu y$ is an antiderivative of $\\mu q$. An antiderivative is fixed by its value at one point, and this value is $y_0$.\n",
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