5. Find L{f(t)}a. f(t) = sin²(t) + tb. f(t) = n² – 8e-2t + (et + 1)2c. f(t) = F, sin yt -cos /v2t , Fo,Y E R4COS2d. f(t)t7/2=

Answered: 5. Find L{f(t)} a. f(t) = sin?(t) + t…

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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5. Find L{f(t)}
a. f(t) = sin²(t) + t
b. f(t) = n² – 8e-
2t + (et + 1)2
c. f(t) = F, sin yt -cos /v2t , Fo,Y E R
4
COS
2
d. f(t)
t7/2
=

Expert Solution

Step 1

As per our limit we can solve only first three subdivisions if you want the answer for other subdivision kindly repost the question and mention the subdivision to be solved.

$a) f (t) = \sin^{2} (t) + t ℒ \{\sin^{2} (t) + t\} = ℒ \{\sin^{2} (t)\} + ℒ \{t\} W e k n o w t h a t \cos 2 t = 1 - 2 \sin^{2} t w h i c h i m p l i e s \sin^{2} t = \frac{1 - \cos 2 t}{2}, ℒ \{\sin^{2} (t) + t\} = ℒ \{\frac{1}{2} - \frac{\cos 2 t}{2}\} + ℒ \{t\} ℒ \{\sin^{2} (t) + t\} = ℒ \{\frac{1}{2}\} - ℒ \{\frac{\cos 2 t}{2}\} + \frac{1}{s^{2}} ℒ \{\sin^{2} (t) + t\} = \frac{1}{2 s} - \frac{1}{2} \cdot \frac{s}{s^{2} + 4} + \frac{1}{s^{2}} ℒ \{\sin^{2} (t) + t\} = \frac{s (s^{2} + 4) - s^{3} + 2 (s^{2} + 4)}{2 s^{2} (s^{2} + 4)} ℒ \{\sin^{2} (t) + t\} = \frac{s^{3} + 4 s - s^{3} + 2 s^{2} + 8}{2 s^{2} (s^{2} + 4)} ℒ \{\sin^{2} (t) + t\} = \frac{2 s^{2} + 4 s + 8}{2 s^{2} (s^{2} + 4)} ℒ \{\sin^{2} (t) + t\} = \frac{s^{2} + 2 s + 4}{s^{4} + 4 s^{2}}$

Step 2

$b) f (t) = π^{2} - 8 e^{- 2 t} + {(e^{t} + 1)}^{2} ℒ \{π^{2} - 8 e^{- 2 t} + {(e^{t} + 1)}^{2}\} = ℒ \{π^{2}\} - ℒ \{8 e^{- 2 t}\} + ℒ \{{(e^{t} + 1)}^{2}\} ℒ \{π^{2} - 8 e^{- 2 t} + {(e^{t} + 1)}^{2}\} = \frac{π^{2}}{s} - 8 ℒ \{e^{- 2 t}\} + ℒ \{e^{2 t} + 2 e^{t} + 1\} ℒ \{π^{2} - 8 e^{- 2 t} + {(e^{t} + 1)}^{2}\} = \frac{π^{2}}{s} - \frac{8}{s + 2} + ℒ \{e^{2 t}\} + 2 ℒ \{e^{t}\} + ℒ \{1\} ℒ \{π^{2} - 8 e^{- 2 t} + {(e^{t} + 1)}^{2}\} = \frac{π^{2}}{s} - \frac{8}{s + 2} + \frac{1}{s - 2} + \frac{2}{s - 1} + \frac{1}{s}$

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