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Find the solution of the given initial value problem. y(t) = = 42() [1 − 2e-(1-2) + e-(-2)] +5e² - 20-² 30) = 21-2e-0-2) +e-20-2)] + 5e²¹ - 5e-2 y(t) y(t) = 21-2-01-2) +-20-2)] + 15e +10e-2 y(t) y(t) = y" + 3y + 2y = u₂(1), y(0) = 5, y/(0) = 5 1₂ (1) 2 42 (1) 2 [1 + 2e-2) +e-2-2)] + 15e¹-10-2 [1-22 +22] + 15e-10e²

Find the solution of the given initial value problem. y(t) = = 42() [1 − 2e-(1-2) + e-(-2)] +5e² - 20-² 30) = 21-2e-0-2) +e-20-2)] + 5e²¹ - 5e-2 y(t) y(t) = 21-2-01-2) +-20-2)] + 15e +10e-2 y(t) y(t) = y" + 3y + 2y = u₂(1), y(0) = 5, y/(0) = 5 1₂ (1) 2 42 (1) 2 [1 + 2e-2) +e-2-2)] + 15e¹-10-2 [1-22 +22] + 15e-10e²

Algebra and Trigonometry (6th Edition)
Algebra and Trigonometry (6th Edition)
6th Edition
ISBN:9780134463216
Author:Robert F. Blitzer
Publisher:Robert F. Blitzer
ChapterP: Prerequisites: Fundamental Concepts Of Algebra
Section: Chapter Questions
Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...
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Find the solution of the given initial value problem. y" + 3y + 2y = ₂(1), y(0) = 5, y(0) = 5 X(t) = [1-2-(1-2) +e-2(1-2)] +5e²¹ - 20-2 21-2e-01-2) + e-21-2)] + 5e²¹ - 5e-² y(t) = y(t) = [1-2e-0-2) +e-20-2)] + 15e +10e-2 [1 + 2e-0¹-2) + e^2(1-2)] + 15e-10e-2 2 ₂ (1) 2 y(t) (1) x(t) = 22 [1-2-0¹-²) +-2-2)] + 15e" - 10e-² Show Transcribed Text the answer is not choice d.
Transcribed Image Text:Find the solution of the given initial value problem. y" + 3y + 2y = ₂(1), y(0) = 5, y(0) = 5 X(t) = [1-2-(1-2) +e-2(1-2)] +5e²¹ - 20-2 21-2e-01-2) + e-21-2)] + 5e²¹ - 5e-² y(t) = y(t) = [1-2e-0-2) +e-20-2)] + 15e +10e-2 [1 + 2e-0¹-2) + e^2(1-2)] + 15e-10e-2 2 ₂ (1) 2 y(t) (1) x(t) = 22 [1-2-0¹-²) +-2-2)] + 15e" - 10e-² Show Transcribed Text the answer is not choice d.
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