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dy 2 + y = 0, y(0) = -3 dt

dy 2 + y = 0, y(0) = -3 dt

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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Do problems 32, 35, 36, 42
39. 2y" + Зу" — 3у' — 2у — е ", У(O) — 0, у'(0) — 0, y"(0) = 1 - 40. у" + 2y" — у' — 2у 3D sin 3t, у(0) — 0, у'() — 0, y"(0) = 1 The inverse forms of the results are S - a L- (s — а)? + b? = ea' cos bt L-1. (s eat sin bt. a)? + b? ] In Problems 41 and 42 use the Laplace transform and these inverses to solve the given initial-value problem. -3t 41. у' + у 3 е cos 2r, y(0) — 0 42. у" — 2у' + 5у %3D 0, у(0) %3 1, у'(0) %3 3
Transcribed Image Text:39. 2y" + Зу" — 3у' — 2у — е ", У(O) — 0, у'(0) — 0, y"(0) = 1 - 40. у" + 2y" — у' — 2у 3D sin 3t, у(0) — 0, у'() — 0, y"(0) = 1 The inverse forms of the results are S - a L- (s — а)? + b? = ea' cos bt L-1. (s eat sin bt. a)? + b? ] In Problems 41 and 42 use the Laplace transform and these inverses to solve the given initial-value problem. -3t 41. у' + у 3 е cos 2r, y(0) — 0 42. у" — 2у' + 5у %3D 0, у(0) %3 1, у'(0) %3 3
2 TRANSFORMS OF DERIVATIVES In Problems 31–40 use the Laplace transform to solve the given initial-value problem. dy 31. dt у %3D 1, у(0) — 0 dy 32. 2 + у %3D 0, у(0) dt | 33. y' + бу %3Dett, y(0) 3D 2 34. y' — у 3D 2 cos 5t, y(0) %3D 0 35. y" + 5у' + 4у %3D 0, у(0) — 1, у'(0) 3D 0 36. y" – 4y' 3 безг — Зе ", у0) — 1, у'(0) — —1 Vz sin V2r, 37. у" + у %3D у(0) %3D 10, у'(0) — 0 38. у" + 9у %3D е, у(0) 3D 0, у'(0) 3D 0
Transcribed Image Text:2 TRANSFORMS OF DERIVATIVES In Problems 31–40 use the Laplace transform to solve the given initial-value problem. dy 31. dt у %3D 1, у(0) — 0 dy 32. 2 + у %3D 0, у(0) dt | 33. y' + бу %3Dett, y(0) 3D 2 34. y' — у 3D 2 cos 5t, y(0) %3D 0 35. y" + 5у' + 4у %3D 0, у(0) — 1, у'(0) 3D 0 36. y" – 4y' 3 безг — Зе ", у0) — 1, у'(0) — —1 Vz sin V2r, 37. у" + у %3D у(0) %3D 10, у'(0) — 0 38. у" + 9у %3D е, у(0) 3D 0, у'(0) 3D 0
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