Exercises

(1)
\(\displaystyle \mathcal {L}(t^3e^{-2t})=\ldots \)
(a)
\(\dfrac {6}{s^4}+2\)
(b)
\(\dfrac {6}{(s+2)^4}\)
(c)
\(\dfrac {6}{(s-2)^4}\)
(d)
\(-\dfrac {2}{s^4+1}\)
(e)
\(\dfrac {6}{s^3(s+2)}\)
(f)
None of these
(2)
Find the value of \(\mathcal {L}((t-1)^2e^{-t})\) at \(s=1\).
(3)
Let \(F(s)=\mathcal {L}(f)\) where \(f(t)=\left \{ \begin{array}{lr} 4-t^2 & \mbox { if }0\leq t<2\\ \\ 0 & \mbox { if }t\geq 2 \end{array}\right .\) Determine \(F(3)\) to the nearest thousandth.
(4)
Suppose \(f(t)\) is piecewise continuous on \([0,\infty )\) and is of exponential order \(s_0\geq 0\). For \(s>s_0\), let \(\displaystyle F(s)=\mathcal {L}(f(t))=\int _0^{\infty } f(t)e^{-st}\,\mathrm {d}t\). Which of the following is \(\displaystyle \frac {dF}{ds}\) for \(s>s_0\)?

Hint: If there exists \(G(s,t)\) with continuous 2nd order partial derivatives for \(s>s_0\) and \(t\geq 0\) satisfying \(G_t=f(t)e^{-st}\) for \(t\geq 0\) then for \(T>0\), \(\displaystyle \int _0^{T} f(t)e^{-st}\,\mathrm {d}t=G(s,T)-G(s,0)\) and thus
\(\displaystyle \dfrac {d}{ds}\displaystyle \int _0^{T} f(t)e^{-st}\,\mathrm {d}t=G_s(s,T)-G_s(s,0)=\int _0^T G_{st}\,\mathrm {d}t=\int _0^T G_{ts}\,\mathrm {d}t\).

(a)
\(\displaystyle s\mathcal {L}(tf(t))\)
(b)
\(\displaystyle -s\mathcal {L}(tf(t))\)
(c)
\(\displaystyle \mathcal {L}(tf(t))\)
(d)
\(\displaystyle -\mathcal {L}(tf(t))\)
(e)
\(\displaystyle -\mathcal {L}(f(t))\)
(f)
None of these
(5)
Use the result of the previous exercise, and that \(\displaystyle \mathcal {L}\left (\sin {(\omega t)}\right )=\frac {\omega }{s^2+\omega ^2}\), to determine the value of the Laplace transform of \(t\sin {(2t)}\) at \(s=1\).