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5.2: Перетворення Лапласа
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$\begin{align*} \mathscr{L}(B \textbf{x}+\textbf{g}) &= \mathscr{L}(B \textbf{x})+\mathscr{L}(\textbf{g}) \\[4pt] &= B \mathscr{L}(\textbf{x})+\mathscr{L}(\textbf{g}) \end{align*}$ \[\begin{align} ... $\begin{align*} \mathscr{L}(B \textbf{x}+\textbf{g}) &= \mathscr{L}(B \textbf{x})+\mathscr{L}(\textbf{g}) \\[4pt] &= B \mathscr{L}(\textbf{x})+\mathscr{L}(\textbf{g}) \end{align*}$ $\begin{align} \mathscr{L} \left(\frac{d \textbf{x}}{dt}\right) &= \int_{0}^{\infty} e^{-(st)} \frac{d \textbf{x}(t)}{dt} dt \\[4pt] &= \textbf{x}(t) \left. \[s \mathscr{L} (\textbf{x})- \textbf{x}(0) = B \mathscr{L}(\textbf{x})+\mathscr{L}(\textbf{g}) \nonumber$
8.2: Резольвент
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$\frac{1}{s-b} = \frac{\frac{1}{s}}{1-\frac{b}{s}} = \frac{1}{s}+\frac{b}{s^2}+ \cdots +\frac{b^{n-1}}{s^n}+\frac{b^n}{s^n} \frac{1}{s-b} \nonumber$ \[(sI-B)^{-1} = s^{-1} \left(I-\frac{B}{s}\right)... $\frac{1}{s-b} = \frac{\frac{1}{s}}{1-\frac{b}{s}} = \frac{1}{s}+\frac{b}{s^2}+ \cdots +\frac{b^{n-1}}{s^n}+\frac{b^n}{s^n} \frac{1}{s-b} \nonumber$ $(sI-B)^{-1} = s^{-1} \left(I-\frac{B}{s}\right)^{-1} \left(\frac{1}{s}+\frac{B}{s^2}+ \cdots +\frac{B^{n-1}}{s^n}+\frac{B^n}{s^n}(sI-B)^{-1}\right) \nonumber$ $(s_{2}I-B)^{-1}-(s_{1}I-B)^{-1} = (s_{2}I-B)^{-1}(s_{1}I-B-s_{2}I+B)(s_{1}I-B)^{-1} \nonumber$