rt By: Search Rank Page Order 01-exam... solutions 9.. solutions._... solutions_... exam22Fe... answers 2... Answers.. 08-exam exam. 1 If f(t) te' H(t-2) for every t≥0, and if C(f) denotes its Laplace transform, then: (A) L(f)(s) = 22-28 (8-1) for s > 1 XL(f)(s) = e²- =2-228-1 for 1 (s-1)2 1 (C) L(f)(s) = for s > 1 (8-1)2 (D) none of the other answers is correct L[t et H(t-2)](3)=d[t-2+2) et 2+2 H(t-2)] (5) = SHIFT = e²d [ (t-2) e²-2 H(t-2)] (s) + 2e² [et-2 H(t-2)] (s) = property -25 -25 = e²es [te] (s) + 2e²e's L[et] (3) = = = e2-25 (- d[e³] (s)) + 2-25 +2e 5-4 2-25 1 2-25 +2e =e 2-25 25-1 = e2-25 (d) =e 2-25 ds 1+25-2-e (S-1) S-1 derivative with respect to s for 571 (5-1)2 (S-1) for 571 + S-I => auswer B } E D

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 92E
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Could you explain how this was solved, I don’t understand the explanation before the use of the shift property As well as the simplification afterwards
rt By:
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01-exam...
solutions 9..
solutions._...
solutions_...
exam22Fe...
answers 2...
Answers..
08-exam
exam.
1 If f(t) te' H(t-2) for every t≥0, and if C(f) denotes its Laplace transform, then:
(A) L(f)(s) =
22-28
(8-1)
for s > 1
XL(f)(s) = e²-
=2-228-1
for 1
(s-1)2
1
(C) L(f)(s) =
for s > 1
(8-1)2
(D) none of the other answers is correct
L[t et H(t-2)](3)=d[t-2+2) et 2+2 H(t-2)] (5) =
SHIFT
= e²d [ (t-2) e²-2 H(t-2)] (s) + 2e² [et-2 H(t-2)] (s) = property
-25
-25
= e²es [te] (s) + 2e²e's L[et] (3) =
=
=
e2-25 (- d[e³] (s)) +
2-25
+2e
5-4
2-25 1
2-25
+2e
=e
2-25
25-1
=
e2-25 (d)
=e
2-25
ds
1+25-2-e
(S-1)
S-1
derivative with respect to s
for 571
(5-1)2
(S-1)
for 571
+
S-I
=> auswer B
}
E
D
Transcribed Image Text:rt By: Search Rank Page Order 01-exam... solutions 9.. solutions._... solutions_... exam22Fe... answers 2... Answers.. 08-exam exam. 1 If f(t) te' H(t-2) for every t≥0, and if C(f) denotes its Laplace transform, then: (A) L(f)(s) = 22-28 (8-1) for s > 1 XL(f)(s) = e²- =2-228-1 for 1 (s-1)2 1 (C) L(f)(s) = for s > 1 (8-1)2 (D) none of the other answers is correct L[t et H(t-2)](3)=d[t-2+2) et 2+2 H(t-2)] (5) = SHIFT = e²d [ (t-2) e²-2 H(t-2)] (s) + 2e² [et-2 H(t-2)] (s) = property -25 -25 = e²es [te] (s) + 2e²e's L[et] (3) = = = e2-25 (- d[e³] (s)) + 2-25 +2e 5-4 2-25 1 2-25 +2e =e 2-25 25-1 = e2-25 (d) =e 2-25 ds 1+25-2-e (S-1) S-1 derivative with respect to s for 571 (5-1)2 (S-1) for 571 + S-I => auswer B } E D
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