-4y"" + 5y" - 5y' − 3y = 0 with y(0) = -3, y'(0) = 3, y"(0) = 2. Taking the Laplace transform and solving for L(y) yields: -8s² + 3s +12 -45³+ 5s²-5s-3 Consider the differential equation L(y) F(s) where F(s) = Note: don't forget to substitute your initial conditions.
-4y"" + 5y" - 5y' − 3y = 0 with y(0) = -3, y'(0) = 3, y"(0) = 2. Taking the Laplace transform and solving for L(y) yields: -8s² + 3s +12 -45³+ 5s²-5s-3 Consider the differential equation L(y) F(s) where F(s) = Note: don't forget to substitute your initial conditions.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
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Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Transcribed Image Text:Consider the differential equation -4y"" + 5y" — 5y' − 3y = 0 with y(0) = −3, y'(0) = 3, y″(0) :
L(y) = F(s) where F(s) =
-8s + 3s + 12
3
2
48° +5s
5s - 3
Note: don't forget to substitute your initial conditions.
= 2. Taking the Laplace transform and solving for L(y) yields:
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Transcribed Image Text:Consider the differential equation −4y" + 5y" — 5y' – 3y = 0 with y(0) = −3, y'(0) = 3, y″(0) = 2. Taking the Laplace transform and solving for L(y) yields:
12s² - 27s+22
-4s³ +5s²-5s-3
L(y) = F(s) where F(s) =
=
Note: don't forget to substitute your initial conditions.
Solution
by Bartleby ExpertFollow-up Question
![The answer above is NOT correct.
L(y) = F(s) where F(s)
-
Entered
[12* (s^2)-27*s+7]/[-4*(s^3)+5*(s^2)-5*s-3]
Note: don't forget to substitute your initial conditions.
Answer Preview
Consider the differential equation -4y" + 5y" — 5y' – 3y = 0 with y(0) = −3, y'(0) = 3, y'(0) = 2. Taking the Laplace transform and solving for L(y) yields:
12s²_ - 27s + 7
-45³ +5s²-5s-3
128² - 27s + 7
-4s³ + 5s² - 5s - 3](https://content.bartleby.com/qna-images/question/008f31aa-ffde-472d-81b9-49fb75f4a0c2/dd1264ba-e4f2-4e36-aaca-23d08390532a/x0cye4d_processed.png)
Transcribed Image Text:The answer above is NOT correct.
L(y) = F(s) where F(s)
-
Entered
[12* (s^2)-27*s+7]/[-4*(s^3)+5*(s^2)-5*s-3]
Note: don't forget to substitute your initial conditions.
Answer Preview
Consider the differential equation -4y" + 5y" — 5y' – 3y = 0 with y(0) = −3, y'(0) = 3, y'(0) = 2. Taking the Laplace transform and solving for L(y) yields:
12s²_ - 27s + 7
-45³ +5s²-5s-3
128² - 27s + 7
-4s³ + 5s² - 5s - 3
Solution
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