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  "body": "\n## Example\n\nLet $f$ and $g$ be continuously differentiable functions on an open interval $J$.\n*d'Alembert's equation*, also called Lagrange's equation, is the implicit first-order equation\n$$y = x\\, f(y') + g(y'),$$\nfor an unknown $y$ of $x$ with $y' \\in J$.\nWhen $f(p) = p$ for all $p$, it is [[Clairaut's equation|0T6D4PE51C]].\nMethod: write $p = y'$ and differentiate, which gives $p = f(p) + \\left(x f'(p) + g'(p)\\right) \\frac{dp}{dx}$.\nWhere $p \\ne f(p)$, view $x$ as a function of $p$; then $(p - f(p))\\, \\frac{dx}{dp} = f'(p)\\, x + g'(p)$ is a linear equation for $x(p)$.\nEach root $p_0$ of $p = f(p)$ gives the line $y = p_0 x + g(p_0)$, which may be a [[singular solution|0CZA0NCFWP]].\n\nInstance: $y = x\\, (y')^2$, so $f(p) = p^2$ and $g = 0$.\nFor $p \\ne 0, 1$, the linear equation is $(p - p^2)\\, dx/dp = 2 p x$, so $dx/dp = 2x/(1 - p)$ and $x = C/(1 - p)^2$.\nThis leads to the solutions $y = (\\sqrt{x} - k)^2$ for $x \u003e 0$ and a constant $k$.\nCheck: $y' = 1 - k/\\sqrt{x}$, so $x (y')^2 = (\\sqrt{x} - k)^2 = y$.\nThe roots $p = 0$ and $p = 1$ of $p = p^2$ give the lines $y = 0$ and $y = x$, which also solve the equation: $x \\cdot 0^2 = 0$ and $x \\cdot 1^2 = x$.\n\n## Explanations\n\n### L1 human\n\nd'Alembert's equation gives $y$ as a combination of $x$ and the slope $y'$, rather than giving the slope directly.\nThe trick is to treat the slope as the new variable: then $x$ satisfies a linear equation, which can be solved.\nBesides the [[general solution|0SKYFQAFYD]] there can be special straight-line solutions, as in Clairaut's equation.\n",
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