(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
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
![(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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcfe11947-ed05-4061-9fa2-a7f116314aff%2Fffbc2b73-61bc-4a92-b596-5e14b40c11d5%2Fwalwmle_processed.jpeg&w=3840&q=75)
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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