Repetitorium HM III

Lösung.

Es ist $\mbox{$F(s):={\operatorname{\mathcal{L}}}(1)(s) = \int_0^\infty\exp(-st)\,dt = \frac{1}{s}$}$ . Damit erhält man
$\mbox{$\displaystyle {\operatorname{\mathcal{L}}}(1+t+t^2)(s) \; =\; F(s) - F'(s) + F''(s) \; =\; s^{-1} + s^{-2} + 2s^{-3}\; . $}$
Es wird
$\mbox{$\displaystyle {\operatorname{\mathcal{L}}}(\exp(3t+1))(s) \; =\; \exp... ...=\; \exp(1){\operatorname{\mathcal{L}}}(1)(s-3) \; =\; \frac{\exp(1)}{s-3}. $}$
Mit
$\mbox{$\displaystyle \begin{array}{rcl} {\operatorname{\mathcal{L}}}(\cos(t... ...athrm{i}s - 1})\vspace*{2mm} \\ & = & \frac{s}{1 + s^2} \\ \end{array} $}$
wird
$\mbox{$\displaystyle \begin{array}{rcl} {\operatorname{\mathcal{L}}}(\int_0... ...s)\vspace*{2mm} \\ & = & \frac{s^2 - 1}{s(1+s^2)^2}\; . \\ \end{array} $}$