Example 5. Find the singular solutions of the differential equation

$$
\left(y^{\prime}\right)^{2}(2-3 y)^{2}=4(1-y)
$$
Solution. Let's find the PDK. Excluding $y^{\prime}$ from the system of equations

$$
\left\{\begin{aligned}
\left(y^{\prime}\right)^{2}(2-3 y)^{2}-4(1-y) & =0 \\
y^{\prime}(2-3 y)^{2} & =0
\end{aligned}\right.
$$

we obtain

$$
(2-3 y)^{2}(1-y)=0 .
$$

Transforming equation (25) to the form

$$
\frac{d x}{d y}= \pm \frac{2-3 y}{2 \sqrt{1-y}}
$$

we find its general integral

$$
y^{2}(1-y)=(x-C)^{2}
$$

Let's find the CDK. Excluding $C$ from the system of equations

$$
\left\{\begin{array}{r}
y^{2}(1-y)-(x-C)^{2}=0 \\
2(x-C)=0
\end{array}\right.
$$

we will have

$$
y^{2}(1-y)=0
$$

Thus, from (26) and (27) we have

$$
\begin{aligned}
& \text { PDK } \equiv(1-y)(2-3 y)^{2}=0 \\
& \text { CDK } \equiv(1-y) y^{2}=0
\end{aligned}
$$

The factor $1-y$ appears in the $p$-discriminant and in the $C$-discriminant to the first degree and gives the envelope, i.e., the function $y=1$ is a special solution of the differential equation (25). By direct substitution, we verify that $y=1$ indeed satisfies the equation.

The equation $2-3 y=0$, appearing in the $p$-discriminant to the second degree and not appearing in the $C$-discriminant at all, gives the place of tangency points ( $\Pi^{2}$ ).

Finally, the equation $y=0$, appearing in the $C$-discriminant to the second degree and not appearing in the $p$-discriminant at all, gives the place of nodal points $\left(P^{2}\right)$ (Fig. 22).

![](https://cdn.mathpix.com/cropped/2024_05_22_a0612bb33256580e6dcdg-065.jpg?height=407&width=708&top_left_y=1113&top_left_x=338)

Fig. 22