Example 1. Find the singular solutions of the differential equation

$$
x y^{\prime}+\left(y^{\prime}\right)^{2}-y=0
$$
## Solution.

a) We find the $p$-discriminant curve. In this case,

$$
F\left(x, y, y^{\prime}\right) \equiv x y^{\prime}+\left(y^{\prime}\right)^{2}-y
$$

and condition (2) takes the form

$$
\frac{\partial F}{\partial y^{\prime}} \equiv x+2 y^{\prime}=0
$$

from which $y^{\prime}=-\frac{x}{2}$. Substituting this expression for $y^{\prime}$ into equation (4), we get

$$
y=-\frac{x^{2}}{4}
$$

Curve (5) is the $p$-discriminant curve of equation (4): it consists of one branch - a parabola.

b) We check whether the $p$-discriminant curve is a solution to the given equation. Substituting (5) and its derivative into (4), we verify that $y=-\frac{x^{2}}{4}$ is a solution to equation (4).

c) We check whether the solution (5) is a singular solution to equation (4). For this, we find the general solution to equation (4). Rewrite (4) as $y=x y^{\prime}+\left(y^{\prime}\right)^{2}$. This is a Clairaut equation. Its general solution is

$$
y=C x+C^{2}
$$

We write the condition for the tangency of two curves $y=y_{1}(x)$ and $y=y_{2}(x)$ at the point with abscissa $x=x_{0}$:

$$
y_{1}\left(x_{0}\right)=y_{2}\left(x_{0}\right), \quad y_{1}^{\prime}\left(x_{0}\right)=y_{2}^{\prime}\left(x_{0}\right)
$$

The first equality expresses the coincidence of the ordinates of the curves, and the second expresses the coincidence of the slopes of the tangents to these curves at the point with abscissa $x=x_{0}$.

Assuming $y_{1}(x)=-\frac{x^{2}}{4}, y_{2}(x)=C x+C^{2}$, we find that conditions (7) take the form

$$
-\frac{x_{0}^{2}}{4}=C x_{0}+C^{2}, \quad-\frac{x_{0}}{2}=C
$$

Substituting $C=-\frac{x_{0}}{2}$ into the first of equations (8), we get

$$
-\frac{x_{0}^{2}}{4}=-\frac{x_{0}^{2}}{2}+\frac{x_{0}^{2}}{4} \text { or }-\frac{x_{0}^{2}}{4}=-\frac{x_{0}^{2}}{4},
$$

i.e., when $C=-\frac{x_{0}}{2}$, the first equation is identically satisfied, since $x_{0}$ is the abscissa of an arbitrary point.

Thus, at each point of curve (5), it is tangent to some other curve of the family (6), namely, the one for which $C=-\frac{x_{0}}{2}$. Therefore, $y=-\frac{x^{2}}{4}$ is a singular solution to equation (4).

d) Geometric interpretation.

The general solution to equation (4) is a family of straight lines (6), and the singular solution (5) is the envelope of this family of straight lines (Fig. 19).

## The envelope of a family of curves

$$
\Phi(x, y, C)=0
$$

is a curve that, at each of its points, is tangent to some curve of the family (9) and each segment of which is tangent to an infinite number of curves from (9).

If (9) is the general integral of equation (1), then the envelope of the family of curves (9), if it exists, will be a special integral curve of this equation. Indeed, at the points of the envelope, the values of $x, y, y^{\prime}$ coincide with the values of $x, y, y^{\prime}$ for the integral curve tangent to the envelope at the point $(x, y)$, and, consequently, at each point of the envelope, the values of $x$, $y, y^{\prime}$ satisfy the equation $F\left(x, y, y^{\prime}\right)=0$, i.e., the envelope is an integral curve.

Furthermore, at each point of the envelope, uniqueness is violated, as at least two integral curves pass through the points of the envelope in the same direction: the envelope itself and the integral curve of the family (9) tangent to it at the considered point. Therefore, the envelope is a special integral curve.

From the course of mathematical analysis, it is known that the envelope is part of the $C$-discriminant curve (briefly, CDC), defined by the system of equations

$$
\left\{\begin{array}{l}
\Phi(x, y, C)=0 \\
\frac{\partial \Phi(x, y, C)}{\partial C}=0
\end{array}\right.
$$

Some branch of the CDC will certainly be an envelope if on it:

1) there exist bounded partial derivatives

$$
\left|\frac{\partial \Phi}{\partial x}\right| \leqslant M, \quad\left|\frac{\partial \Phi}{\partial y}\right| \leqslant N
$$

where $M$ and $N$ are constants;
2)

$$
\frac{\partial \Phi}{\partial x} \neq 0 \quad \text { or } \quad \frac{\partial \Phi}{\partial y} \neq 0
$$

Remark. Conditions 1) and 2) are only sufficient, and therefore branches of the CDC on which one of these conditions is violated can also be envelopes.