d'Alembert's equation

Example

Let $f$ and $g$ be continuously differentiable functions on an open interval $J$. d'Alembert's equation, also called Lagrange's equation, is the implicit first-order equation $$y = x\, f(y') + g(y'),$$ for an unknown $y$ of $x$ with $y' \in J$. When $f(p) = p$ for all $p$, it is Clairaut's equation. Method: write $p = y'$ and differentiate, which gives $p = f(p) + \left(x f'(p) + g'(p)\right) \frac{dp}{dx}$. Where $p \ne f(p)$, view $x$ as a function of $p$; then $(p - f(p))\, \frac{dx}{dp} = f'(p)\, x + g'(p)$ is a linear equation for $x(p)$. Each root $p_0$ of $p = f(p)$ gives the line $y = p_0 x + g(p_0)$, which may be a singular solution.

Instance: $y = x\, (y')^2$, so $f(p) = p^2$ and $g = 0$. For $p \ne 0, 1$, the linear equation is $(p - p^2)\, dx/dp = 2 p x$, so $dx/dp = 2x/(1 - p)$ and $x = C/(1 - p)^2$. This leads to the solutions $y = (\sqrt{x} - k)^2$ for $x > 0$ and a constant $k$. Check: $y' = 1 - k/\sqrt{x}$, so $x (y')^2 = (\sqrt{x} - k)^2 = y$. The 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 $x \cdot 1^2 = x$.