-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.

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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:
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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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.
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.
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Follow-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
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
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