Clairaut's equation: y = xp + f(p) where p = $\frac{dy}{dx}$. Differentiating: p = p + xp' + f'(p)p', so p'(x + f'(p)) = 0. Either p' = 0 (giving p = c, so y = cx + f(c), the general solution — a family of lines) or x + f'(p) = 0 (giving the singular solution, an envelope). The general solution is obtained by replacing p with the arbitrary constant c. The singular solution is found by eliminating p between y = xp + f(p) and x = -f'(p). Example: y = xp + 1/p. General solution: y = cx + 1/c. Singular solution: from x = 1/$p^2$, p = 1/sqrt(x), y = $\frac{x}{sqrt}$(x) + sqrt(x) = 2*sqrt(x), i.e., $y^2$ = 4x.