Example 6. Integrate the equation $y=2 x y^{\prime}+\ln y^{\prime}$.
Solution. Let $y^{\prime}=p$, then $y=2 x p+\ln p$. Differentiating, we find

$$
p d x=2 p d x+2 x d p+\frac{d p}{p}
$$

from which $p \frac{d x}{d p}=-2 x-\frac{1}{p}$ or $\frac{d x}{d p}=-\frac{2}{p} x-\frac{1}{p^{2}}$.

We have obtained a first-order equation, linear in $x$; solving it, we find

$$
x=\frac{C}{p^{2}}-\frac{1}{p}
$$

Substituting the found value of $x$ into the expression for $y$, we finally get

$$
x=\frac{C}{p^{2}}-\frac{1}{p}, \quad y=\ln p+\frac{2 C}{p}-2
$$

The Clairaut equation has the form

$$
y=x y^{\prime}+\psi\left(y^{\prime}\right)
$$

The method of solution is the same as for the Lagrange equation. The general solution of the Clairaut equation has the form

$$
y=C x+\psi(C)
$$

The Clairaut equation may also have a singular solution, which is obtained by eliminating $p$ from the equations $y=x p+\psi(p), x+\psi^{\prime}(p)=0$.