Example 3. Solve the system of equations

$$
\left\{\begin{array}{l}
t \frac{d x}{d t}=-x+y t \\
t^{2} \frac{d y}{d t}=-2 x+y t
\end{array}\right.
$$
Solution. From the first equation of the system, we find

$$
y=\frac{x}{t}+\frac{d x}{d t}, \quad \text { so that } \quad \frac{d y}{d t}=-\frac{x}{t^{2}}+\frac{1}{t} \cdot \frac{d x}{d t}+\frac{d^{2} x}{d t^{2}} \text {. }
$$

Substituting these expressions for $y$ and $\frac{d y}{d t}$ into the second equation, we get

$$
t^{2} \frac{d^{2} x}{d t^{2}}+t \frac{d x}{d t}-x=-2 x+x+t \frac{d x}{d t}, \quad \text { or } \quad t^{2} \frac{d^{2} x}{d t^{2}}=0
$$

Assuming $t \neq 0$, from the last equation we have $\frac{d^{2} x}{d t^{2}}=0$ and after integration we get $x=C_{1}+C_{2} t$. Now it is easy to find

$$
y=\frac{x}{t}+\frac{d x}{d t}=\frac{C_{1}+C_{2} t}{t}+C_{2}=2 C_{2}+\frac{C_{1}}{t}
$$

The general solution of the given system is

$$
x=C_{1}+C_{2} t, \quad y=\frac{C_{1}}{t}+2 C_{2}, \quad t \neq 0
$$

Remark. Not every system of differential equations can be reduced to a single higher-order equation. For example, $\frac{d x}{d t}=x$, $\frac{d y}{d t}=y$ cannot be reduced to a single second-order equation. Its general solution is $x=C_{1} e^{t}, y=C_{2} e^{t}$.

Problems for independent solving

Solve the following systems of differential equations by the method of elimination:

776. $\left\{\begin{array}{l}\frac{d x}{d t}=-9 y, \\ \frac{d y}{d t}=x .\end{array}\right.$
777. 

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

778. $\begin{cases}\frac{d x}{d t}+3 x+4 y=0, & x(0)=1, \\ \frac{d y}{d t}+2 x+5 y=0, & y(0)=4 .\end{cases}$
779. 

$\begin{cases}\frac{d x}{d t}=x+5 y, & x(0)=-2 \\ \frac{d y}{d t}=-x-3 y, & y(0)=1\end{cases}$

780. $\left\{\begin{array}{l}4 \frac{d x}{d t}-\frac{d y}{d t}+3 x=\sin t \\ \frac{d x}{d t}+y=\cos t\end{array}\right.$
781. $\left\{\begin{array}{l}\frac{d x}{d t}=-y+z, \\ \frac{d y}{d t}=z, \\ \frac{d z}{d t}=-x+z .\end{array}\right.$
782. $\left\{\begin{array}{l}\frac{d x}{d t}=-y+z, \\ \frac{d y}{d t}=z, \\ \frac{d z}{d t}=-x+z .\end{array}\right.$
783. $\left\{\begin{array}{l}\frac{d x}{d t}=y+z, \\ \frac{d y}{d t}=x+z, \\ \frac{d z}{d t}=x+y .\end{array}\right.$
784. $\left\{\begin{array}{l}\frac{d^{2} x}{d t^{2}}=y \\ \frac{d^{2} y}{d t^{2}}=x\end{array}\right.$
785. $\left\{\begin{array}{l}\frac{d^{2} x}{d t^{2}}+\frac{d y}{d t}+x=0, \\ \frac{d x}{d t}+\frac{d^{2} y}{d t^{2}}=0 .\end{array}\right.$
786. $\left\{\begin{array}{l}\frac{d^{2} x}{d t^{2}}=3 x+y \\ \frac{d y}{d t}=-2 x\end{array}\right.$
787. $\left\{\begin{array}{l}\frac{d^{2} x}{d t^{2}}=x^{2}+y, \quad x(0)=x^{\prime}(0)=1, \\ \frac{d y}{d t}=-2 x \frac{d x}{d t}+x, \quad y(0)=0 .\end{array}\right.$

## § 21. Finding Integrable Combinations. Symmetric Form of a System of Differential Equations

$1^{\circ}$. Finding Integrable Combinations. This method of integrating a system of differential equations

$$
\frac{d x_{k}}{d t}=f_{k}\left(t, x_{1}, x_{2}, \ldots, x_{n}\right), \quad k=1,2, \ldots, n
$$

consists of the following: using arithmetic operations (addition, subtraction, multiplication, division) on the equations of the system (1), we form so-called integrable combinations, i.e., equations of the form

$$
F\left(t, u, \frac{d u}{d t}\right)=0
$$

where $u$ is some function of the unknown functions $x_{1}(t), x_{2}(t), \ldots, x_{n}(t)$. Each integrable combination provides one first integral. If $n$ independent first integrals of the system (1) are found, then its integration is complete; if $m$ independent first integrals are found, where $m<n$, then the system (1) is reduced to a system with a smaller number of unknown functions.