The roots of a d.e are -1, 2+√3 What is the solution? y = ex(c₁ cos 3x + c₂ sin 3x) + c3 cOS X + C4 sin x y=c₁e-x + c₂e(2+√3)x+ cze(2-√3)x -X -X y = C₁e-x + C₂ cos 2x + c3 sin 2x y = ₁е-x + c₂e²+√³x + C₂e²+√3x -X

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.2: Trigonometric Equations
Problem 69E
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The roots of a d.e are
-1, 2+√3
What is the solution?
y = ex(c₁ cos 3x + c₂ sin 3x) + c3 cos x + C4 sin x
-X
y = c₁e-x + c₂e(²+√3)x+ C₂e(2-√3)x
-X
y = C₁е-x + C₂ cos 2x + c3 sin 2x
-X
y = C₁e¯x + c₂e²+√√³x + C3e²+√³x
Transcribed Image Text:The roots of a d.e are -1, 2+√3 What is the solution? y = ex(c₁ cos 3x + c₂ sin 3x) + c3 cos x + C4 sin x -X y = c₁e-x + c₂e(²+√3)x+ C₂e(2-√3)x -X y = C₁е-x + C₂ cos 2x + c3 sin 2x -X y = C₁e¯x + c₂e²+√√³x + C3e²+√³x
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