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  "body": "\n## Proof\n\nWe show that $W$ solves $W' = -p W$ and then solve this linear equation.\n\n1. As solutions of the [[homogeneous equation|0EQZSAB9EC]], $y_1$ and $y_2$ are twice differentiable on $I$. So $y_1, y_2, y_1', y_2'$ are differentiable.\n2. By the [[product rule|0V9A43H7WV]] and [[linearity of differentiation|0M0TKFRX2A]], the [[Wronskian|0N5NF1RGBP]] $W = y_1 y_2' - y_1' y_2$ is differentiable on $I$, with $W' = y_1' y_2' + y_1 y_2'' - y_1'' y_2 - y_1' y_2' = y_1 y_2'' - y_1'' y_2$.\n3. The equation gives $y_k'' = -p\\, y_k' - q\\, y_k$ for $k = 1, 2$. Substituting, $W' = y_1 \\left(-p y_2' - q y_2\\right) - \\left(-p y_1' - q y_1\\right) y_2$.\n4. The terms with $q$ cancel, and $W' = -p\\left(y_1 y_2' - y_1' y_2\\right) = -p W$ on $I$.\n5. Let $P(x) = \\int_{x_0}^{x} p(t)\\, dt$. Since $p$ is continuous, [[$P$ is differentiable with $P' = p$|0NC2A3HWGT]]. Also $P(x_0) = 0$.\n6. Let $G = W e^{P}$. By the product rule, the [[chain rule|0RXX2PT1MQ]] and [[the derivative of the exponential|0EX9GFJ0B3]], $G' = W' e^{P} + W p\\, e^{P} = (-p W + p W) e^{P} = 0$.\n7. So [[$G$ is constant|003QDZ4T16]] on $I$, and $G(x) = G(x_0) = W(x_0) e^{0} = W(x_0)$ for each $x \\in I$.\n8. By the [[exponential of a sum|0E8HNVBCSP]], $e^{P(x)} e^{-P(x)} = e^{0} = 1$. Multiplying $W(x) e^{P(x)} = W(x_0)$ by $e^{-P(x)}$ gives $W(x) = W(x_0) \\exp\\left(-\\int_{x_0}^{x} p(t)\\, dt\\right)$.\n\n## Idea\n\nDifferentiating $W$ leaves only second derivatives. The equation expresses them through first derivatives and the function, and the result is $-p W$. A linear first-order equation $W' = -p W$ has exponential solutions, fixed by the value at $x_0$.\n",
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