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7. y"+y'=e¹ +38(t-6), y(0) = -1, y'(0) = 4

7. y"+y'=e¹ +38(t-6), y(0) = -1, y'(0) = 4

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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In Exercises 1-20 solve the initial value problem. Where indicated by C/G, graph the solution. y" + 3y + 2y = 6e²t + 28(t − 1), y(0) = 2, y'(0) = −6 -6 1. 2. C/Gy' + y' - 2y = −10e-t +58(t-1), 3. y4y = 2e¯t +58(t−1), y(0) = −1, C/Gy" + y = sin 3t+26(t — π/2), y" + 4y = 4 + 8(t - 3π), y(0) = 0, yy=8+28(t− 2), y(0) = −1, 4. 5. 6. 7. y"+y' = et + 38(t− 6), 8. 9. y(0) = −1, y(0) = 7, y'(0) = −9 y'(0) = 2 y(0) = 1, y'(0) = −1 y'(0) = 1 y'(0) = 1 y'(0) = 4 y" + 4y = 8e²t + 8(t-π/2), y(0) = 8, y'(0) = 0 C/G y + 3y + 2y = 1+8(t−1), y(0) = 1, y'(0) = −1
Transcribed Image Text:In Exercises 1-20 solve the initial value problem. Where indicated by C/G, graph the solution. y" + 3y + 2y = 6e²t + 28(t − 1), y(0) = 2, y'(0) = −6 -6 1. 2. C/Gy' + y' - 2y = −10e-t +58(t-1), 3. y4y = 2e¯t +58(t−1), y(0) = −1, C/Gy" + y = sin 3t+26(t — π/2), y" + 4y = 4 + 8(t - 3π), y(0) = 0, yy=8+28(t− 2), y(0) = −1, 4. 5. 6. 7. y"+y' = et + 38(t− 6), 8. 9. y(0) = −1, y(0) = 7, y'(0) = −9 y'(0) = 2 y(0) = 1, y'(0) = −1 y'(0) = 1 y'(0) = 1 y'(0) = 4 y" + 4y = 8e²t + 8(t-π/2), y(0) = 8, y'(0) = 0 C/G y + 3y + 2y = 1+8(t−1), y(0) = 1, y'(0) = −1
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