Example 3. Find a particular solution of the system

$$
\left\{\begin{array}{l}
\frac{d x}{d t}=1-\frac{1}{y} \\
\frac{d y}{d t}=\frac{1}{x-t}
\end{array}\right.
$$

satisfying the initial conditions $\left.x\right|_{t=0}=1,\left.y\right|_{t=0}=1$.
Solution. Let's write the given system in the form

$$
\left\{\begin{array} { c } 
{ y ( \frac { d x } { d t } - 1 ) = - 1 , } \\
{ ( x - t ) \frac { d y } { d t } = 1 , }
\end{array} \quad \text { or } \quad \left\{\begin{array}{c}
y \frac{d(x-t)}{d t}=-1 \\
(x-t) \frac{d y}{d t}=1
\end{array}\right.\right.
$$

Adding the last equations term by term, we get

$$
y \frac{d(x-t)}{d t}+(x-t) \frac{d y}{d t}=0, \quad \text { or } \quad \frac{d}{d t}[(x-t) y]=0 .
$$

From this, we find the first integral $(x-t) y=C_{1}$. Since $x-t=\frac{C_{1}}{y}$, the second equation of the system will take the form $\frac{d y}{d t}=\frac{y}{C_{1}}$, from which $y=C_{2} e^{t / C_{1}}$. Thus,

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

from which we obtain the general solution

$$
x=t+\frac{C_{1}}{C_{2}} e^{-t / C_{1}}, \quad y=C_{2} e^{t / C_{1}}
$$

Setting $t=0$ in these equations, we find $1=\frac{C_{1}}{C_{2}}, 1=C_{2}$, i.e., $C_{1}=C_{2}=1$, and the desired particular solution is

$$
x=t+e^{-t}, \quad y=e^{t}
$$