Answered: DEFINITION 10.1.1 Let f be a function…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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use (10.1.1) to determine L[f].

Q. f(t)=sinh bt, whereb is constant.

DEFINITION 10.1.1
Let f be a function defined on an interval [0, o0). The Laplace transform of f is
the function F(s) defined by
F(0) = e"fmdt,
(10.1.1)
%3D
provided that the improper integral converges. We will usually denote the Laplace
transform of f by L[f].

Expert Solution

Step 1

Given: $f (t) = \sin h b t$
To find: The Laplace transform of $f (t), L [f]$ .

Step 2

We know that the Laplace transform of $f$ is the function $F (s)$ defined by-
$F (s) = \int_{0}^{\infty} e^{- s t} f (t) d t - 1$
Here,
$f (t) = \sin h b t$
This can be expressed as : $f (t) = [\frac{e^{b t} - e^{- b t}}{2}]$
$∴ F (s) = \int_{0}^{\infty} e^{- s t} [\frac{e^{b t} - e^{- b t}}{2}] d t = \frac{1}{2} \int_{0}^{\infty} e^{- s t} (e^{b t}) d t - \frac{1}{2} \int_{0}^{\infty} e^{- s t} (e^{- b t}) d t (linear combination of Laplace transform) = \frac{1}{2} L [e^{b t}] - \frac{1}{2} L [e^{- b t}] [Using 1] = \frac{1}{2} (\frac{1}{s - b}) - \frac{1}{2} (\frac{1}{s + b}) [∵ L [e^{\pm a t}] = \frac{1}{s \mp a}] = \frac{1}{2} (\frac{1}{s - b} - \frac{1}{s + b}) = \frac{1}{2} (\frac{s + b - (s - b)}{(s + b) (s - b)}) = \frac{1}{2} (\frac{2 b}{s^{2} - b^{2}}) = \frac{b}{s^{2} - b^{2}} \Rightarrow L [f] = \frac{b}{s^{2} - b^{2}}$

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