Question

Compute the Laplace transform. Your answer should be a function of the variable \( s \) : \[ \mathcal{L}\left\{2+u_{3 / 2}(t) e^{2 t} \cos (\pi t)\right\}= \] \( \square \) You may find the following formulas useful: \[ \begin{aligned} \cos (b t+\pi) & =-\cos (b t) \\ \sin (b t+\pi) & =-\sin (b t) \\ \cos \left(b t+\frac{\pi}{2}\right) & =-\sin (b t) \\ \sin \left(b t+\frac{\pi}{2}\right) & =\cos (b t) \end{aligned} \]

Name: compute the laplace transform your answer should be a function of the variable s mathcallleft2u3 2t e2 t cos pi tright square you may find the following formulas useful beginaligned cos b tpi cos b 2
Uploaded: 2024-11-02T21:30:24-08:00
Duration: 3 min 13 s
Channel: Andreas Papavassiliou
Description: compute the laplace transform your answer should be a function of the variable s mathcallleft2u3 2t e2 t cos pi tright square you may find the following formulas useful beginaligned cos b tpi cos b 2

          Compute the Laplace transform. Your answer should be a function of the variable \( s \) :
\[
\mathcal{L}\left\{2+u_{3 / 2}(t) e^{2 t} \cos (\pi t)\right\}=
\]
\( \square \)

You may find the following formulas useful:
\[
\begin{aligned}
\cos (b t+\pi) & =-\cos (b t) \\
\sin (b t+\pi) & =-\sin (b t) \\
\cos \left(b t+\frac{\pi}{2}\right) & =-\sin (b t) \\
\sin \left(b t+\frac{\pi}{2}\right) & =\cos (b t)
\end{aligned}
\]

Added by User U.

Differential Equations and Linear Algebra

Stephen W. Goode, Scott A. Annin 4th Edition

Chapter 10

Instant Answer

Solved by Expert Andreas Papavassiliou

11/02/2024

Step 1

The given function is \( f(t) = 2 + u_{3/2}(t) e^{2t} \cos(\pi t) \), where \( u_{3/2}(t) \) is the unit step function that shifts the function by \( 3/2 \). Show more…

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Compute the Laplace transform. Your answer should be a function of the variable \( s \) : \[ \mathcal{L}\left\{2+u_{3 / 2}(t) e^{2 t} \cos (\pi t)\right\}= \] \( \square \) You may find the following formulas useful: \[ \begin{aligned} \cos (b t+\pi) & =-\cos (b t) \\ \sin (b t+\pi) & =-\sin (b t) \\ \cos \left(b t+\frac{\pi}{2}\right) & =-\sin (b t) \\ \sin \left(b t+\frac{\pi}{2}\right) & =\cos (b t) \end{aligned} \]