Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
Related questions
Question
could you explain why the 2s+3 becomes 2s+1+2? and the how the inverse laplace transform was taken step by step?
![Calculation:
Rewrite the given function.
Y [s]
=
Now,
=
2s+3
s²+s+1
2s+1+2
s²+8+1/+²³/
2s+1+2
s²+2·s · 1/2 + ( - ) ² + ³
2s+1+2
(s+ 12 ) ² + ( +/- ) ²
2s+1
(s++)² + (+)²
+
*
(+)
(s + 1 ) ² + ( √/²³ ) ²
√3
2(s+ -)
+
(s + 1 ) ² + ( √/3³ ) ² (s + 12 ) ² + ( √/3³ ) ²
L-¹ [Y (s)] = L-¹
+
2(s+1)
[ (s+ + ) ² + ( + ³ ) ²
= 2e-¹¹ cos (t) + 4e¯¹ sin(t)
-t
+ L-1
ਜਾਂਦਾ
(s+ + ) ² + ( + ) ²
2s+3
Hence L-¹ [243] = 2e-¹ cos (√t) + ¹ sin(t).
t
s²+s+1](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F652e5396-12b6-4e5e-865e-eecd6b9e3fee%2F91d8d8ce-1e4d-411c-a32f-3524a8fdc379%2Fn0m737u_processed.png&w=3840&q=75)
Transcribed Image Text:Calculation:
Rewrite the given function.
Y [s]
=
Now,
=
2s+3
s²+s+1
2s+1+2
s²+8+1/+²³/
2s+1+2
s²+2·s · 1/2 + ( - ) ² + ³
2s+1+2
(s+ 12 ) ² + ( +/- ) ²
2s+1
(s++)² + (+)²
+
*
(+)
(s + 1 ) ² + ( √/²³ ) ²
√3
2(s+ -)
+
(s + 1 ) ² + ( √/3³ ) ² (s + 12 ) ² + ( √/3³ ) ²
L-¹ [Y (s)] = L-¹
+
2(s+1)
[ (s+ + ) ² + ( + ³ ) ²
= 2e-¹¹ cos (t) + 4e¯¹ sin(t)
-t
+ L-1
ਜਾਂਦਾ
(s+ + ) ² + ( + ) ²
2s+3
Hence L-¹ [243] = 2e-¹ cos (√t) + ¹ sin(t).
t
s²+s+1
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