Example 4. Solve the system

$$
\left\{\begin{array}{l}
\frac{d x}{d t}=x+2 y+16 t e^{t} \\
\frac{d y}{d t}=2 x-2 y
\end{array}\right.
$$
Solution. The characteristic equation

$$
\left|\begin{array}{cc}
1-\lambda & 2 \\
2 & -2-\lambda
\end{array}\right|=0, \text { or } \quad \lambda^{2}+\lambda-6=0
$$

The roots of the characteristic equation are $\lambda_{1}=2, \lambda_{2}=-3$. The general solution of the homogeneous system corresponding to system (15) is:

$$
\tilde{x}=2 C_{1} e^{2 t}+C_{2} e^{-3 t}, \quad \tilde{y}=C_{1} e^{2 t}-2 C_{2} e^{-3 t}
$$

We seek a particular solution of the non-homogeneous system of equations (15) in the form

$$
x_{\text {p }}=(A t+B) e^{t}, \quad y_{\text {p }}=(M t+N) e^{t}
$$

Substituting (16) into (15) and canceling $e^{t}$, we get:

$$
\begin{aligned}
A t+B+A & =A t+B+2 M t+2 N+16 t \\
M t+N+M & =2 A t+2 B-2 M t-2 N
\end{aligned}
$$

Thus, $A=-12, B=-13, M=-8, N=-6$. Therefore,

$$
x_{\text {p }}=-(12 t+13) e^{t}, \quad y_{\text {p }}=-(8 t+6) e^{t}
$$

The general solution of the original system is:

$$
\begin{aligned}
& x=\tilde{x}+x_{p}=2 C_{1} e^{2 t}+C_{2} e^{-3 t}-(12 t+13) e^{t} \\
& y=\tilde{y}+y_{p}=C_{1} e^{2 t}-2 C_{2} e^{-3 t}-(8 t+6) e^{t}
\end{aligned}
$$

## Problems for Independent Solution

Integrate non-homogeneous linear systems with constant coefficients.

817. $\left\{\begin{array}{l}\frac{d x}{d t}=y, \\ \frac{d y}{d t}=1-x .\end{array}\right.$
818. $\left\{\begin{array}{l}\frac{d x}{d t}=-y+\sin t, \\ \frac{d y}{d t}=x+\cos t .\end{array}\right.$
819. $\left\{\begin{array}{l}\frac{d x}{d t}=3-2 y, \\ \frac{d y}{d t}=2 x-2 t .\end{array}\right.$
820. $\left\{\begin{array}{l}\frac{d x}{d t}=x+y+e^{t}, \\ \frac{d y}{d t}=x+y-e^{t} .\end{array}\right.$
821. $\left\{\begin{array}{l}\frac{d x}{d t}=4 x-5 y+4 t-1, x(0)=0, \\ \frac{d y}{d t}=x-2 y+t, y(0)=0 .\end{array}\right.$
822. $\begin{cases}\frac{d x}{d t}=y-x+e^{t}, & x(0)=1, \\ \frac{d y}{d t}=x-y+e^{t}, & y(0)=1 .\end{cases}$
823. $\left\{\begin{array}{l}\frac{d x}{d t}+y=t^{2}, \\ \frac{d y}{d t}-x=t .\end{array}\right.$
824. $\left\{\begin{array}{l}\frac{d x}{d t}+\frac{d y}{d t}+y=e^{-t}, \\ 2 \frac{d x}{d t}+\frac{d y}{d t}+2 y=\sin t\end{array}\right.$
825. $\left\{\begin{array}{l}\frac{d x}{d t}=2 x+y-2 z+2-t, \\ \frac{d y}{d t}=1-x, \\ \frac{d z}{d t}=x+y-z+1-t .\end{array}\right.$
826. $\left\{\begin{array}{l}\frac{d x}{d t}+x+2 y=2 e^{-t}, \quad x(0)=1, \\ \frac{d y}{d t}+y+z=1, \quad y(0)=1, \\ \frac{d z}{d t}+z=1, \quad z(0)=1 .\end{array}\right.$

$3^{\circ}$. Construction of Integrable Combinations (D'Alembert's Method). This method is used to construct integrable combinations when solving systems of linear equations with constant coefficients. Let's show its application for solving systems of two equations:

$$
\left\{\begin{array}{l}
\frac{d x}{d t}=a_{1} x+b_{1} y+f_{1}(t) \\
\frac{d y}{d t}=a_{2} x+b_{2} y+f_{2}(t)
\end{array}\right.
$$

Multiply the second equation by some number $\lambda$ and add it term by term to the first equation:

$$
\frac{d(x+\lambda y)}{d t}=\left(a_{1}+\lambda a_{2}\right) x+\left(b_{1}+\lambda b_{2}\right) y+f_{1}(t)+\lambda f_{2}(t)
$$

Rewrite the last equation as

$$
\frac{d(x+\lambda y)}{d t}=\left(a_{1}+\lambda a_{2}\right)\left(x+\frac{b_{1}+\lambda b_{2}}{a_{1}+\lambda a_{2}} y\right)+f_{1}(t)+\lambda f_{2}(t)
$$

Choose the number $\lambda$ such that

$$
\frac{b_{1}+\lambda b_{2}}{a_{1}+\lambda a_{2}}=\lambda
$$

Then (18) reduces to an equation linear in $x+\lambda y$

$$
\frac{d(x+\lambda y)}{d t}=\left(a_{1}+\lambda a_{2}\right)(x+\lambda y)+f_{1}(t)+\lambda f_{2}(t)
$$

Integrating which, we get

$$
x+\lambda y=e^{\left(a_{1}+\lambda a_{2}\right) t}\left\{C+\int\left[f_{1}(t)+\lambda f_{2}(t)\right] e^{-\left(a_{1}+\lambda a_{2}\right) t} d t\right\}
$$

If equation (19) has distinct real roots $\lambda_{1}$ and $\lambda_{2}$, then from (20) we obtain two first integrals of system (17), and thus the integration of this system is complete.