Example 1. Solve the system

$$
\left\{\begin{array}{l}
\frac{d x_{1}}{d t}=2\left(x_{1}^{2}+x_{2}^{2}\right) t \\
\frac{d x_{2}}{d t}=4 x_{1} x_{2} t
\end{array}\right.
$$
Solution. Adding the two equations term by term, we get

$$
\frac{d\left(x_{1}+x_{2}\right)}{d t}=\left(x_{1}+x_{2}\right)^{2} 2 t
$$

from which

$$
-\frac{1}{x_{1}+x_{2}}=t^{2}-C_{1}, \text { or } \frac{1}{x_{1}+x_{2}}+t^{2}=C_{1} \text {. }
$$

Subtracting the two equations term by term, we get

$$
\frac{d\left(x_{1}-x_{2}\right)}{d t}=2 t\left(x_{1}-x_{2}\right)^{2}
$$

from which

$$
\frac{1}{x_{1}-x_{2}}+t^{2}=C_{1}
$$

Thus, two first integrals of the given system are found

$$
\begin{aligned}
& \psi_{1}\left(t, x_{1}, x_{2}\right)=t^{2}+\frac{1}{x_{1}+x_{2}}=C_{1} \\
& \psi_{2}\left(t, x_{1}, x_{2}\right)=t^{2}+\frac{1}{x_{1}-x_{2}}=C_{2}
\end{aligned}
$$

which are independent, since the Jacobian is not zero:

$$
\frac{D\left(\psi_{1}, \psi_{2}\right)}{D\left(x_{1}, x_{2}\right)}=\left|\begin{array}{ll}
\frac{\partial \psi_{1}}{\partial x_{1}} & \frac{\partial \psi_{1}}{\partial x_{2}} \\
\frac{\partial \psi_{2}}{\partial x_{1}} & \frac{\partial \psi_{2}}{\partial x_{2}}
\end{array}\right|=\left|\begin{array}{cc}
-\frac{1}{\left(x_{1}+x_{2}\right)^{2}} & -\frac{1}{\left(x_{1}+x_{2}\right)^{2}} \\
-\frac{1}{\left(x_{1}-x_{2}\right)^{2}} & -\frac{1}{\left(x_{1}-x_{2}\right)^{2}}
\end{array}\right|=-\frac{2}{\left(x_{1}^{2}-x_{2}^{2}\right)^{2}} \neq 0
$$

The general integral of the system (2) is

$$
t^{2}+\frac{1}{x_{1}+x_{2}}=C_{1}, \quad t^{2}+\frac{1}{x_{1}-x_{2}}=C_{2}
$$

Solving the system (3) for the unknown functions, we get the general solution of the system (2):

$$
x_{1}=\frac{C_{1}+C_{2}-2 t^{2}}{2\left(C_{1}-t^{2}\right)\left(C_{2}-t^{2}\right)}, \quad x_{2}=\frac{C_{2}-C_{1}}{2\left(C_{1}-t^{2}\right)\left(C_{2}-t^{2}\right)} .
$$