Find the solution of the given initial value problem. y" + y = g(t), y(0) = 5, y' (0) = ? where 0si<6 g(1) | 3, t > 6 1 1 y(t) = ;0 = – - cost) – u6(1)[(t – 6) – cos(t – 6)] + 5sint + 7cost 1 y(1) = (t – sint) - ju6(1)[(t – 6) – sin(t – 6)] + 5cost + 7sint y(t) = (t – sint) – u6 (t)[(t – 6) – sin(t – 6)] + 5cost + 7sint 1 1 y(t) = ¿(t – sint) - ju6(1)[(t – 6) – sin(t – 6)] + 7cost + 5sint 1 y(t) = (t – sint) +ju6(t)[(t – 6) – sin(t – 6)] + 5cost + 7sint
Find the solution of the given initial value problem. y" + y = g(t), y(0) = 5, y' (0) = ? where 0si<6 g(1) | 3, t > 6 1 1 y(t) = ;0 = – - cost) – u6(1)[(t – 6) – cos(t – 6)] + 5sint + 7cost 1 y(1) = (t – sint) - ju6(1)[(t – 6) – sin(t – 6)] + 5cost + 7sint y(t) = (t – sint) – u6 (t)[(t – 6) – sin(t – 6)] + 5cost + 7sint 1 1 y(t) = ¿(t – sint) - ju6(1)[(t – 6) – sin(t – 6)] + 7cost + 5sint 1 y(t) = (t – sint) +ju6(t)[(t – 6) – sin(t – 6)] + 5cost + 7sint
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![Find the solution of the given initial value problem.
y" +y = g(t), y(0) = 5, y'(0) = 7
where
(늘0Si<6
0<1 <
g(t) :
3, t 2 6
Y() = -
(t – cost) – u6(1)[(t – 6) – cos(t – 6)] + 5sint + 7cost
1
:- sint) – 46 (t)[(t – 6) – sin(t – 6)] + 5cost + 7sint
y(t)
O y(t) = (1 – sinf) – us(1)[(t – 6) – sin(t – 6)] + 5cost + 7sint
y(t):
:- sint) – u6 (1)[I(± – 6) – sin(t – 6)] + 7cost + 5sint
= « -
1
y(t)
:- sint) + u6 (t)[(t – 6) – sin(t – 6)] + 5cost + 7sint](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F77cac1a6-5ad9-4f72-bdd9-21a202e53df4%2Fbc5630b0-a5fc-45f6-8f38-fd8907b73d66%2F7jgvxdq_processed.png&w=3840&q=75)
Transcribed Image Text:Find the solution of the given initial value problem.
y" +y = g(t), y(0) = 5, y'(0) = 7
where
(늘0Si<6
0<1 <
g(t) :
3, t 2 6
Y() = -
(t – cost) – u6(1)[(t – 6) – cos(t – 6)] + 5sint + 7cost
1
:- sint) – 46 (t)[(t – 6) – sin(t – 6)] + 5cost + 7sint
y(t)
O y(t) = (1 – sinf) – us(1)[(t – 6) – sin(t – 6)] + 5cost + 7sint
y(t):
:- sint) – u6 (1)[I(± – 6) – sin(t – 6)] + 7cost + 5sint
= « -
1
y(t)
:- sint) + u6 (t)[(t – 6) – sin(t – 6)] + 5cost + 7sint
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