Example 1. Solve the equation $y^{\prime \prime}+y^{\prime}-2 y=x^{3}-2$.
Solution. We have: $k^{2}+k-2=0, k_{1}=-2, k_{2}=1, y_{00}=C_{1} e^{-2 x}+$ $+C_{2} e^{x}$. The right-hand side of the non-homogeneous equation is $P_{n}(x) e^{x}=$ $=\left(x^{3}-2\right) e^{0 x}$, i.e., $n=3, \lambda=0$ is not a root of the characteristic equation. According to point 1) of theorem 3, a particular solution of the non-homogeneous equation should be chosen in the form: $y_{4}=A x^{3}+B x^{2}+C x+D$, where $A, B, C$ and $D$ are unknown numbers. To determine them, we substitute the expressions for $y_{4}, y_{4}^{\prime}, y_{4}^{\prime \prime}$ into the original equation.

We organize the calculations as follows. We write the expressions for $y_{4}, y_{4}^{\prime}$, $y_{4}^{\prime \prime}$ in a column, and before them, the corresponding coefficients of the equation.

$$
\begin{aligned}
& -2: y_{\mathrm{4}}=A x^{3}+B x^{2}+C x+D \\
& 1: y_{4}^{\prime}=3 A x^{2}+2 B x+C \\
& 1: y_{4}^{\prime \prime}=6 A x+2 B
\end{aligned}
$$

We obtained the system of equations by equating the coefficients of the same powers of the variable $x$ in the right and left parts of the equation, obtained by substituting the expressions for $y_{4}, y_{4}^{\prime}, y_{4}^{\prime \prime}$ into it (method of undetermined coefficients):

$$
\begin{aligned}
& x^{3}:-2 A=1, \quad A=-\frac{1}{2} ; \\
& x^{2}:-2 B+3 A=0, \quad B=\frac{3 A}{2}=-\frac{3}{4} \\
& x^{1}:-2 C+2 B+6 A=0, \quad C=B+3 A=-\frac{3}{4}-\frac{3}{2}=-\frac{9}{4} \\
& x^{0}:-2 D+C+2 B=-2, \quad D=\frac{2+C+2 B}{2}=-\frac{7}{8}
\end{aligned}
$$

The general solution of the non-homogeneous equation ${ }^{2} y_{\text {on }}$ is composed according to theorem 1.

Answer. $y_{\text {on }}=y_{\text {oo }}+y_{4}=C_{1} e^{-2 x}+C_{2} e^{x}-\frac{x^{3}}{2}-\frac{3 x^{2}}{4}-\frac{9 x}{4}-\frac{7}{8}$.