Example 6. Find the general solution of the system of equations

$$
\frac{d t}{2 x}=\frac{d x}{-\ln t}=\frac{d y}{\ln t-2 x}
$$
Solution. The first integrable combination is $\frac{d t}{2 x}=-\frac{d x}{\ln t}$. Separating variables and integrating, we find the first integral

$$
t(\ln t-1)+x^{2}=C_{1} .
$$

The second integrable combination is obtained using the derivatives of the proportion (19). For this, we add the numerators and denominators of the fractions of system (20):

$$
\frac{d t}{2 x}=\frac{d x}{-\ln t}=\frac{d y}{\ln t-2 x}=\frac{d t+d x+d y}{0}
$$

here $\lambda_{1}=1, \lambda_{2}=1, \lambda_{3}=1$. From this, $d t+d x+d y=0$, or $d(t+x+y)=0$ and, therefore,

$$
t+x+y=C_{2} .
$$

The first integrals (21) and (22) give the general integral of system (20)

$$
x^{2}+t(\ln t-1)=C_{1}, \quad x+y+t=C_{2}
$$

from which we find the general solution of the system

$$
x= \pm \sqrt{C_{1}+t(\ln t-1)}, \quad y=C_{2}-t \mp \sqrt{C_{1}+t(\ln t-1)}
$$