Use the accompanying tables of Laplace transforms and properties of Laplace transforms to find the Laplace transform of the function below. Note that an appropriate trigonometric identity may be nece 3t sin 27t Properties of Laplace Transforms Click here to view th Click here to view th i Table of Laplace Transforms L{3t sin 27t} =O L{f+g} = L{f} + L{g} f(t) F(s) = L{f}{s) L{cf} = cf{f} for any constant c 1 s>0 L{e atr1)}(s) = L{}(s - a) L {t'>(s) = sL{f}(s) – f(0) L{t">(s) = s² L{f}{s) – sf(0) – f'(0) L{m}(s) = s^ L{{s) – s^ - 1f(0) – s^- 2f(0) – ..- - fn – 1)(0) 1 e at 1 S-a s>0 n! n,n =1,2,.. sn+1 s>0 L {t^{()}(s) = ( – 1)n(L{0(s)) dsn sin bt Print Done cos bt s>0 Enter your answer in n! e atn̟ n =1,2,.. (s-an+1s>a

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
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Chapter7: Distance And Approximation
Section7.5: Applications
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Use the accompanying tables of Laplace transforms and properties of Laplace transforms to find the Laplace transform of the function below. Note that an appropriate trigonometric identity may be neces
3t sin 27t
Properties of Laplace Transforms
Click here to view th
Click here to view th
Table of Laplace Transforms
L{3t sin 27t} =|
L{f+g} = L{f} + L{g}
f(t)
F(s) = L{f{s)
L{cf} = cI{f} for any constant c
1
L{e atr)} (s) = L{f}{s- a)
L {t'>(s) = sI{f}(s) – f(0)
L {t'">(s) = s?L{f(s) – sf(0) – f'(0)
L {fn)}(s) = s"L{f}{s) – s" -1f(0) - sn-2t'(0) – ... -
1
e at
s>0
s-a
f(n - 1) (0)
tn, n =1,2,...
n!
L {t"f(1)}(s) = (– 1)"
dn
-(L{f}{(s))
ds"
b
sin bt
s+ h2. s>0
cos bt
Print
Done
s2+ b2"
s>0
Enter your answer in
n!
e atn, n=1,2,.-.
(s-a)n+1.s>a
Transcribed Image Text:Use the accompanying tables of Laplace transforms and properties of Laplace transforms to find the Laplace transform of the function below. Note that an appropriate trigonometric identity may be neces 3t sin 27t Properties of Laplace Transforms Click here to view th Click here to view th Table of Laplace Transforms L{3t sin 27t} =| L{f+g} = L{f} + L{g} f(t) F(s) = L{f{s) L{cf} = cI{f} for any constant c 1 L{e atr)} (s) = L{f}{s- a) L {t'>(s) = sI{f}(s) – f(0) L {t'">(s) = s?L{f(s) – sf(0) – f'(0) L {fn)}(s) = s"L{f}{s) – s" -1f(0) - sn-2t'(0) – ... - 1 e at s>0 s-a f(n - 1) (0) tn, n =1,2,... n! L {t"f(1)}(s) = (– 1)" dn -(L{f}{(s)) ds" b sin bt s+ h2. s>0 cos bt Print Done s2+ b2" s>0 Enter your answer in n! e atn, n=1,2,.-. (s-a)n+1.s>a
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