4.79. Integrate the Lagrange equation

$$
y=x\left(\frac{d y}{d x}\right)^{2}-\frac{d y}{d x}
$$
The problem is solved. Let's move to the parametric representation $p=\frac{d y}{d x}$, then according to formula (4.7) we get:

$$
\begin{aligned}
& a(p)=p^{2}, \quad b(p)=-p \\
& \frac{d x}{x p}+\frac{2}{p-1} x=\frac{1}{p^{2}-p} \\
& x=u(p) v(p) \\
& v \frac{d u}{d p}+u \frac{d v}{d p} \frac{2}{p-1} u v=\frac{1}{p^{2}-p} \\
& v \frac{d u}{d p}+u\left[\frac{d v}{d p}+\frac{2}{p-1} v\right]=\frac{1}{p^{2}-p} \\
& {\left[\frac{d v}{d p}+\frac{2}{p-1} v\right]=0} \\
& \frac{d v}{v}=-\frac{2 d p}{p-1}, \quad \int \frac{d v}{v}=-2 \int \frac{d p}{p-1} \\
& \ln |v|=-2 \ln |p-1| \\
& v=\frac{1}{(p-1)^{2}}, \\
& \frac{1}{(p-1)^{2}} \frac{d u}{d p}=\frac{1}{p(p-1)} \\
& d u=\frac{(p-1) d p}{p} \\
& \int d u=\int d p-\int \frac{d p}{p} \\
& u=p-\ln |p|+C
\end{aligned}
$$

Thus,

$$
\left\{\begin{array}{l}
x=u v=\frac{p-\ln |p|+C}{(p-1)^{2}} \\
y=x p^{2}-p
\end{array}\right.
$$