Example 2. Find the singular solutions of the differential equation

$$
x\left(y^{\prime}\right)^{2}-2 y y^{\prime}+4 x=0, \quad x>0
$$

knowing its general integral

$$
x^{2}=C(y-C) \text {. }
$$
## Solution.

a) We find the $C$-discriminant curve. We have

$$
\Phi(x, y, C) \equiv C(y-C)-x^{2}
$$

so

$$
\frac{\partial \Phi}{\partial C} \equiv y-2 C
$$

from which $C=\frac{y}{2}$. Substituting this value of $C$ into (14), we get

$$
x^{2}=\frac{y}{2}\left(y-\frac{y}{2}\right)
$$

from which

$$
(y-2 x)(y+2 x)=0, \quad \text { or } \quad y= \pm 2 x .
$$

This is the $C$-discriminant curve: it consists of two lines $y=2 x$ and $y=-2 x$.
b) By direct substitution, we verify that each branch of the CDS is a solution of equation (13).

c) We prove that each of the solutions (15) is a singular solution of equation (13). Indeed, since

$$
\frac{\partial \Phi}{\partial x}=-2 x, \quad \frac{\partial \Phi}{\partial y}=C
$$

on each branch of the CDS we have

$$
\left|\frac{\partial \Phi}{\partial x}\right|=|-2 x| \leqslant 2 b
$$

(assuming that the solution $y(x)$ of equation (13) is considered on the interval $00$, so one of the conditions (12) is satisfied. Therefore, conditions (11) and (12) are met, and thus the lines (15) are the envelopes of the parabolas (14).

Thus, it is established that each of the solutions (15) is a singular solution. $\square$

In the search for singular solutions, the following symbolic schemes prove useful:

$$
\begin{aligned}
& \text { PK } \equiv 0 \cdot 3 \cdot \Pi^{2}=0 \\
& \text { CDS } \equiv 0 \cdot y^{2} \cdot 3^{3}=0
\end{aligned}
$$

Scheme (16) means that the equation of the $p$-discriminant curve can break down into three equations:

1) $\mathrm{O}=0$ - the equation of the envelope;
2) $3=0$ - the equation of the geometric locus of points of inflection (return);
3) $\Pi=0$ - the equation of the geometric locus of points of tangency of integral curves, and the factor $\Pi$ enters the PDC squared.

Scheme (17) means that the equation of the $C$-discriminant curve can break down into three equations:

1) $\mathrm{O}=0-$ the equation of the envelope;
2) $\mathrm{y}=0$ - the equation of the geometric locus of nodal points, and the factor Y enters the CDS squared;
3) $3=0$ - the equation of the geometric locus of points of inflection, and the factor 3 enters the CDS cubed.

It is not necessary that all components of the PDC and CDS appear in the relations (16) and (17) for each problem.

Of all the geometric loci, only the envelope is a singular solution of the differential equation. Finding the envelope is simplified by the fact that it enters the schemes (16) and (17) in the first degree.

Regarding other geometric loci (points of inflection, nodal points, and points of tangency), additional analysis is required in each specific case. The fact that some factor enters the PDC squared (and does not enter the CDS at all) indicates that there may be a geometric locus of points of tangency of integral curves. Similarly, if some factor enters the CDS squared (and does not enter the PDC at all), then there may be a geometric locus of nodal points. Finally, if a factor enters the PDC in the first degree and the CDS in the third, there may be a geometric locus of points of inflection.