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(f) y +2y +10y=eH(t-1), y(0)=-1, y'(0) = 0 (involve partial fraction, completing the square, first shift theorem and second shift theorem; Answer: y(1)=(1-1)-(1-1) cos(31-3)+sin (31-3) ]-e* cos 31 - e* sin 31 13 13 3

(f) y +2y +10y=eH(t-1), y(0)=-1, y'(0) = 0 (involve partial fraction, completing the square, first shift theorem and second shift theorem; Answer: y(1)=(1-1)-(1-1) cos(31-3)+sin (31-3) ]-e* cos 31 - e* sin 31 13 13 3

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section9.10: Partial Fractions
Problem 2E
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(f) y +2y' +10y = e"H(t- 1), y(0) = -1,y'(0) = 0 (involve partial fraction, completing the square, first shift theorem and second shift theorem; Answer: y(1) = (1-1)-e (1-1) cos(31 -3)+sin (31-3) -e" cos31 - sin 31) 3 13 13
Transcribed Image Text:(f) y +2y' +10y = e"H(t- 1), y(0) = -1,y'(0) = 0 (involve partial fraction, completing the square, first shift theorem and second shift theorem; Answer: y(1) = (1-1)-e (1-1) cos(31 -3)+sin (31-3) -e" cos31 - sin 31) 3 13 13
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