Given the differential equation and its initial conditions y" - y' 2y 0 with y(0) = -2 and y'(0) = 5

Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter2: Systems Of Linear Equations
Section2.4: Applications
Problem 27EQ
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q13

- 2S + 3
A Y(s) =
(s - 2) (s + 1)
2S + 7
B Y(s)
(s - 2) (s + 1)
7 - 25
(C)
Y(s):
(S +2)(s - 1)
7 - 25
Y(s)
(s - 2) (s + 1)
(D
5S - 7
E
E Y(s) =
(s - 2)(s + 1)
7
5S
F Y(s)
(s - 2) (s + 1)
Transcribed Image Text:- 2S + 3 A Y(s) = (s - 2) (s + 1) 2S + 7 B Y(s) (s - 2) (s + 1) 7 - 25 (C) Y(s): (S +2)(s - 1) 7 - 25 Y(s) (s - 2) (s + 1) (D 5S - 7 E E Y(s) = (s - 2)(s + 1) 7 5S F Y(s) (s - 2) (s + 1)
Now that we know how to convert individual functions, we can practice converting entire differential equations into Laplace
Transforms. Once everything is converted, there will be no more derivatives left and all we have to do is solve for the "Y(S)" term. In
other words, we have used the Laplace Transforms to convert a calculus problem into an algebra problem.
Given the differential equation and its initial conditions
y" - y' - 2y = 0 with y(0) = -2 and y'(0) = 5
Use the Laplace Transform rules for derivatives to convert this function into F(s) and then solve for Y(s).
L{ y(1)) = Y(s)
L{ y'(1)} = S Y(s) - y(0)
L{ y'(1)) = s2 Y(s) - Sy(0) - y'(0)
Transcribed Image Text:Now that we know how to convert individual functions, we can practice converting entire differential equations into Laplace Transforms. Once everything is converted, there will be no more derivatives left and all we have to do is solve for the "Y(S)" term. In other words, we have used the Laplace Transforms to convert a calculus problem into an algebra problem. Given the differential equation and its initial conditions y" - y' - 2y = 0 with y(0) = -2 and y'(0) = 5 Use the Laplace Transform rules for derivatives to convert this function into F(s) and then solve for Y(s). L{ y(1)) = Y(s) L{ y'(1)} = S Y(s) - y(0) L{ y'(1)) = s2 Y(s) - Sy(0) - y'(0)
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