4.78. Find the singular solutions of the equation

$$
F\left(x, y, \frac{d y}{d x}\right)=\left(\frac{d y}{d x}\right)^{2}-(6 x+y) \frac{d y}{d x}+6 x y=0
$$
Solving this equation with respect to $\frac{d y}{d x}$, we obtain two equations: $\frac{d y_{1}}{d x}=y_{1}$ and $\frac{d y_{2}}{d x}=6 x$, the right-hand sides of which satisfy the condition of existence and uniqueness of the solution to the Cauchy problem at any point in the plane $(x, y)$. The general solutions of these equations are: $y_{1}=C_{1} e^{x}$, $y_{2}=3 x^{2}+C_{2}$, where $C_{1}$ and $C_{2}$ are arbitrary constants of integration. The geometric locus of points where the condition $F_{p}^{\prime}\left(x_{0}, y_{0}, p_{0}\right) \neq 0$ of the theorem on the existence and uniqueness of the solution is violated, are the points of the line $y=-6 x-12$. Indeed,

$$
\left.\frac{\partial F}{\partial y^{\prime}}\right|_{y=6 x}=2 y^{\prime}-6 x-y=-12-6 x-y=-12-6 x+6 x+12=0
$$

The line $y=6 x$ is a $p$-discriminant curve. Excluding $p$ from the equations $F(x, y, p)=p^{2}-(6 x+y) p+6 x y=0$ $\frac{\partial F}{\partial p}=2 p-(6 x+y)=0$, we obtain $\Phi(x, y)=y-6 x=0$, however, $y=6 x$ is not an integral curve of the original equation, therefore, $y=6 x$ is not a special solution.