Example 8. Integrate the system of differential equations $\left\{\begin{array}{l}x^{\prime}=\frac{y+e^{x}}{y+e^{t}}, \\ y^{\prime}=\frac{y^{2}-e^{x+t}}{y+e^{t}} .\end{array}\right.$
Solution. In this system, the unknown functions $x$ and $y$ depend on the argument $t$.

1) Multiply both sides of the second equation by $e^{-t}$:

$$
e^{-t} y^{\prime}=\frac{y^{2} e^{-t}-e^{x}}{y+e^{t}}
$$

Add this equality to the first equation of the system and to the identity $-e^{-t} y=-e^{-t} y$. We get:

$$
e^{-t} y^{\prime}+x^{\prime}-e^{-t} y=\frac{y^{2} e^{-t}+y}{y+e^{t}}-e^{-t} y
$$

Notice that the right-hand side of this equality is zero. The left-hand side, however, is the sum of derivatives: $\left(e^{-t} y\right)^{\prime}+x^{\prime}$. Thus, we have obtained a non-trivial combination: $\left(e^{-t} y\right)^{\prime}+x^{\prime}=0$, from which $e^{-t} y+x=C_{1}$,
which represents part of the general solution of the system (we have only one constant).

2) We will obtain another, similar combination in no less "creative" a way. Multiply both sides of the first equation by $-e^{-x} y$, both sides of the second equation by $e^{-x}$, add the results, and to them, add the equality $t^{\prime}=1$ (the derivative of the argument is one). We have:

$$
-e^{-x} y x^{\prime}+e^{-x} y^{\prime}+t^{\prime}=\frac{-y^{2} e^{-x}-y}{y+e^{t}}+\frac{y^{2} e^{-x}-e^{-t}}{y+e^{t}}+1
$$

The right-hand side is zero, and in the left-hand side $\left(e^{-x} y\right)^{\prime}+t^{\prime}=0$, i.e., $e^{-x} y+t=C_{2}$.

3) The general integral (general solution in implicit form) is obtained from the system of equalities thus obtained

Answer. $\left\{\begin{array}{l}e^{-t} y+x=C_{1}, \\ e^{-x} y+t=C_{2} .\end{array}\right.$