Clairaut's equation

Example

Let $f$ be a twice continuously differentiable function on an open interval $J$. Clairaut's equation is the implicit first-order equation $$y = x\, y' + f(y'),$$ for an unknown $y$ with $y' \in J$. For each constant $C \in J$, the line $y = C x + f(C)$ is a solution. These lines form the general solution. If $f'' \ne 0$ on $J$, the curve with the parametric form $x = -f'(p)$, $y = f(p) - p f'(p)$ for $p \in J$ is also a solution. It is the envelope of the lines, and it is a singular solution.

Remarks

Differentiation of the equation gives $y''\left(x + f'(y')\right) = 0$. The case $y'' = 0$ gives the lines, and the case $x + f'(y') = 0$ gives the envelope. Example: for $f(p) = p^2$, the lines are $y = C x + C^2$, and the singular solution is the parabola $y = -x^2/4$. Each line of the general solution is tangent to the singular solution at one point.

Explanations

L1 human

Clairaut's equation has a family of straight lines as solutions, one for each slope $C$. These lines wrap around a curve, their envelope, and this curve is also a solution. No choice of $C$ gives the envelope, so it is a new, singular solution.