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In Exercises 5–8, show that each function is a solution of the given initial value problem. Differential Initial Solution equation candidate equation 2 5. y' + y = п y(-In 2) = y = e¯*tan¬'(2e*) 1 + 4e2* 6. y' = e* - 2xy y(2) = 0 у %3D (х — 2)е * cos x 7. xy' + y = -sin x, x > 0 2 8. xy' = xy – y², x > 1 y(e) = e In x

In Exercises 5–8, show that each function is a solution of the given initial value problem. Differential Initial Solution equation candidate equation 2 5. y' + y = п y(-In 2) = y = e¯*tan¬'(2e*) 1 + 4e2* 6. y' = e* - 2xy y(2) = 0 у %3D (х — 2)е * cos x 7. xy' + y = -sin x, x > 0 2 8. xy' = xy – y², x > 1 y(e) = e In x

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 5–8, show that each function is a solution of the given initial value problem. Differential Initial Solution equation candidate equation 2 5. y' + y = п y(-In 2) = y = e¯*tan¬'(2e*) 1 + 4e2* 6. y' = e* - 2xy y(2) = 0 у %3D (х — 2)е * cos x 7. xy' + y = -sin x, x > 0 2 8. xy' = xy – y², x > 1 y(e) = e In x
Transcribed Image Text:In Exercises 5–8, show that each function is a solution of the given initial value problem. Differential Initial Solution equation candidate equation 2 5. y' + y = п y(-In 2) = y = e¯*tan¬'(2e*) 1 + 4e2* 6. y' = e* - 2xy y(2) = 0 у %3D (х — 2)е * cos x 7. xy' + y = -sin x, x > 0 2 8. xy' = xy – y², x > 1 y(e) = e In x
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