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Prove that the given equation is a solution to the given DE: 1. %3D Gjезх + Сzе 2х + Сзе*; у" — бу" + 11у'- бу %3D 0 2. = x2 + 4x ; xy' x² + y 3. у %3D ез*; у" — 4у" — 4y' + 16у %3 0

Prove that the given equation is a solution to the given DE: 1. %3D Gjезх + Сzе 2х + Сзе*; у" — бу" + 11у'- бу %3D 0 2. = x2 + 4x ; xy' x² + y 3. у %3D ез*; у" — 4у" — 4y' + 16у %3 0

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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Prove that the given equation is a solution to the given DE: 1. = C,e3x + C2e2x + C3e* ; y" – 6y" + 11y' - 6y = 0 %3D 2. = x2 + 4x ; xy' x² + y 3. у %3Dе3х; у"— 4y" — 4у' + 1бу —D 0 4. y = (x + C)e-*; y'+ y = e-* 5. x = 2t + C; y = Ct + 3 ; 2(y')² – xy' = 3 - y
Transcribed Image Text:Prove that the given equation is a solution to the given DE: 1. = C,e3x + C2e2x + C3e* ; y" – 6y" + 11y' - 6y = 0 %3D 2. = x2 + 4x ; xy' x² + y 3. у %3Dе3х; у"— 4y" — 4у' + 1бу —D 0 4. y = (x + C)e-*; y'+ y = e-* 5. x = 2t + C; y = Ct + 3 ; 2(y')² – xy' = 3 - y
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