Solve h" +4h' + 3h = 8(t), h(0) = 0, h'(0) = 0. Express the forced response of y" + 4y + 3y convolution integral. Use the result in (a) (b) to solve y" + 4y' + 3y = 31 [43] (a) h(t)e ¹ - / e f(t), y(0) 3 0, y'(0) = y(0) = 0, y'(0) = 0. 0, using a (b) y(t) = (h* f)(t) = f h(t – T)f(T)dT 31 (c) y(t) = e 'In(1 + e²) + (−² − 1) e ¹ + ½ e ²¹ - ½ e ³¹ ln(1 + e²) + (-1/2 + 1¹12₂2²) e (− (A hint for evaluating the integral: substitute v = 1+ e¹) ' 2 3
Solve h" +4h' + 3h = 8(t), h(0) = 0, h'(0) = 0. Express the forced response of y" + 4y + 3y convolution integral. Use the result in (a) (b) to solve y" + 4y' + 3y = 31 [43] (a) h(t)e ¹ - / e f(t), y(0) 3 0, y'(0) = y(0) = 0, y'(0) = 0. 0, using a (b) y(t) = (h* f)(t) = f h(t – T)f(T)dT 31 (c) y(t) = e 'In(1 + e²) + (−² − 1) e ¹ + ½ e ²¹ - ½ e ³¹ ln(1 + e²) + (-1/2 + 1¹12₂2²) e (− (A hint for evaluating the integral: substitute v = 1+ e¹) ' 2 3
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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Question
solution provided by instructor for practice, therefore its not graded work
![[43] (a) Solve h" +4h! +3h = 8(t), h(0) = 0, h'(0) = 0.
(b) Express the forced response of y" + 4y' + 3y
convolution integral.
(c) Use the result in (a) (b) to solve y" + 4y+3y=y(0) = 0, y'(0) = 0.
3L
[43] (a) h(t) = ½e ¹ - /e
f(t), y(0) =
0, y'(0)
21
= 0, using a
(b) y(t) = (h* f)(t) = f h(t – T)f(T)dT
e ² + 1/ e
L
(c) y(t) = ½e 'ln(1 + e²) + (−¹m² − 1) e
(A hint for evaluating the integral: substitute v = 1+ e¹)
-½e 3 ln(1 + ²) + (+¹₂²) 6
31
e](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8c89c353-252e-4eca-b673-064be33ffdb0%2F29763e93-d146-4b8b-bb03-a45df0c69424%2F5et4mzd_processed.png&w=3840&q=75)
Transcribed Image Text:[43] (a) Solve h" +4h! +3h = 8(t), h(0) = 0, h'(0) = 0.
(b) Express the forced response of y" + 4y' + 3y
convolution integral.
(c) Use the result in (a) (b) to solve y" + 4y+3y=y(0) = 0, y'(0) = 0.
3L
[43] (a) h(t) = ½e ¹ - /e
f(t), y(0) =
0, y'(0)
21
= 0, using a
(b) y(t) = (h* f)(t) = f h(t – T)f(T)dT
e ² + 1/ e
L
(c) y(t) = ½e 'ln(1 + e²) + (−¹m² − 1) e
(A hint for evaluating the integral: substitute v = 1+ e¹)
-½e 3 ln(1 + ²) + (+¹₂²) 6
31
e
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43b solution provided by instructor for practice, not graded (no solution for b, please solve)
![[43] (a) Solve h" +4h' + 3h = 8(t), h(0) = 0, h'(0) = 0.
(b)
calculate h'(0+), is it h'(0) in the initial condition of this question?
If no, give a reason. What is the role of 8(t) and what is the connection between h'(0+) and 8(t)?
Physical (or other areas) explanations are welcome.
[43] (a) h(t)=e¹-e
31](https://content.bartleby.com/qna-images/question/8c89c353-252e-4eca-b673-064be33ffdb0/065eaf32-9a01-4d59-a824-4556e22fbc43/lpr4mi_processed.png)
Transcribed Image Text:[43] (a) Solve h" +4h' + 3h = 8(t), h(0) = 0, h'(0) = 0.
(b)
calculate h'(0+), is it h'(0) in the initial condition of this question?
If no, give a reason. What is the role of 8(t) and what is the connection between h'(0+) and 8(t)?
Physical (or other areas) explanations are welcome.
[43] (a) h(t)=e¹-e
31
Solution
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