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

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

Calculus: Early Transcendentals
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...
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- 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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