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Put $p=y^{\prime }$ and differentiate the general form of Clairaut's equation:Cancellation givesFor $f(p)=1/(p-1)$ the equation is defined only when $p\ne 1$. There are two types of smooth branch.

First, constant $p=C\ne 1$ gives the straight-line familySecond, $x+f^{\prime }(p)=0$ gives $x=1/(p-1{)}^2$, so $x>0$ and $p=1\pm x^{-1/2}$. Substituting in the original equation gives its two singular envelope branches,Their derivatives are indeed $1\pm 1/\sqrt{x}$, verifying the required value of $p$. There is no finite-slope continuation of either envelope through $x=0$, and no real envelope branch for $x<0$.

For completeness, if “solution” requires only a continuous first derivative, one can join a line from the family to the corresponding envelope at their tangency point $x=(C-1{)}^{-2}$. The function and its first derivative match there, so such piecewise branches also solve the original first-order equation. They are not twice continuously differentiable at a nontrivial join. The lines and envelope branches above exhaust the smooth branches; allowing these tangent joins supplies the additional $C^1$ solutions.