Let $ f(t) $ be a function on $[0, \infty)$. The Laplace transform of $ f $ is the function $ F $ defined by the integral \[F(s) = \int_{0}^{\infty} e^{-st}f(t) \, dt.\]Use this definition to determine the Laplace transform of the following function.\[ f(t) = e^{t} + (t + 3)^{2}\]The Laplace transform of $ f(t) $ is $ F(s) = $ $\boxed{\quad}$

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Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

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Let $ f(t) $ be a function on $[0, \infty)$. The Laplace transform of $ f $ is the function $ F $ defined by the integral

\[
F(s) = \int_{0}^{\infty} e^{-st}f(t) \, dt.
\]

Use this definition to determine the Laplace transform of the following function.

\[
f(t) = e^{t} + (t + 3)^{2}
\]

The Laplace transform of $ f(t) $ is $ F(s) = $ $\boxed{\quad}$

Expert Solution

Step 1

Given that, $f (t) = e^{t} + {(t + 3)}^{2}$

To determine the Laplace transform of $f (t)$ .

The Laplace transform of a function, $f (t)$ is the function, $F$ defined by,

$F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t$

Therefore, the Laplace transform of $f (t)$ is

$\begin{array}{rcl} F (s) & = & \int_{0}^{\infty} e^{- s t} (e^{t} + {(t + 3)}^{2}) d t \\ = & \int_{0}^{\infty} e^{(1 - s) t} d t + \int_{0}^{\infty} {(t + 3)}^{2} e^{- s t} d t \end{array}$

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