11.2.1. Solve the differential equation using the operational method: $x^{\prime \prime}-x^{\prime}-20 \cdot x=\mathrm{e}^{-t}, x(0)=0, x^{\prime}(0)=4$.
Solution. We transition from originals to images of functions.

Let $x(t) \risingdotseq X(p)=x$,

$$
\begin{aligned}
& x^{\prime}(t) \risingdotseq p \cdot X(p)-x_{0}=p \cdot X-0=p \cdot X \\
& x^{\prime \prime}(t) \risingdotseq p^{2} \cdot X(p)-p \cdot x_{0}-x_{0}^{\prime}=p^{2} \cdot X-4 .
\end{aligned}
$$

Then, taking into account that the image of the right-hand side is $e^{-1} \risingdotseq \frac{1}{p+1}$, we obtain the operator equation in the form:

$$
\begin{gathered}
p^{2} \cdot X-p \cdot X-20 \cdot X-4=\frac{1}{p+1} \\
X \cdot\left(p^{2}-p-20\right)=\frac{1}{p+1}+4
\end{gathered}
$$

From this, we find

$$
\begin{gathered}
X=\frac{1}{(p+1)\left(p^{2}-p-20\right)}+\frac{4}{p^{2}-p-20}=\frac{1}{(p+1)(p-5)(p+4)}+\frac{4}{(p-5)(p+4)}= \\
=-\frac{1}{18} \cdot \frac{1}{p+1}+\frac{1}{54} \cdot \frac{1}{p-5}+\frac{1}{27} \cdot \frac{1}{p+4}+\frac{4}{9} \cdot \frac{1}{p-5}-\frac{4}{9} \cdot \frac{1}{p+4}= \\
=-\frac{1}{18} \cdot \frac{1}{p+1}+\frac{2}{54} \cdot \frac{1}{p-5}-\frac{11}{27} \cdot \frac{1}{p+4}
\end{gathered}
$$

Using the table of images, we find the desired particular solution:

$$
x(t)=-\frac{1}{18} \cdot e^{-t}+\frac{25}{54} \cdot e^{5 t}-\frac{11}{27} \cdot e^{-4 t}
$$

## SECTION 6

## PROBABILITY THEORY AND MATHEMATICAL STATISTICS

## 6.1. Basic Concepts and Theorems of Probability Theory