EXAMPLE 12: Using Laplace transforms, find the solution of the initial value problem y" - 4y + 4y = 64 sin 2t y(0) = 0, y'(0) = 1 y" - 4y' + 4y = 64 sin 2t y(0) = 0, y'(0) = 1 Taking Laplace transform of both sides of (1), we have 64×2 5² +4 On putting the value of y(0) and y'(0) in (2), we get s²y-1-4s y+4y= 128 s² +4 Solution: [²y-sy(0)-y()]-4 y-y()]+ 4y = (²-45+4)y=1+ + 128 or (S-2) ÿ=1+ s +4 y=L* | -3 1 128 1 8 16 8s + (S-2) (s-2) (s²+4) (s-2) s-2 (S-2) s² +4 8.s 17 s-2 (S-2) s² +4 + y=-8e²¹ +17e²¹ +8 cos 2t 128 s² +4 Ans. (1) (2)
EXAMPLE 12: Using Laplace transforms, find the solution of the initial value problem y" - 4y + 4y = 64 sin 2t y(0) = 0, y'(0) = 1 y" - 4y' + 4y = 64 sin 2t y(0) = 0, y'(0) = 1 Taking Laplace transform of both sides of (1), we have 64×2 5² +4 On putting the value of y(0) and y'(0) in (2), we get s²y-1-4s y+4y= 128 s² +4 Solution: [²y-sy(0)-y()]-4 y-y()]+ 4y = (²-45+4)y=1+ + 128 or (S-2) ÿ=1+ s +4 y=L* | -3 1 128 1 8 16 8s + (S-2) (s-2) (s²+4) (s-2) s-2 (S-2) s² +4 8.s 17 s-2 (S-2) s² +4 + y=-8e²¹ +17e²¹ +8 cos 2t 128 s² +4 Ans. (1) (2)
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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Kindly solve this partial fraction
How below step comes from
Thank you
![EXAMPLE 12: Using Laplace transforms, find the solution of the
initial value problem
y" - 4y + 4y = 64 sin 2t
y(0) = 0, y' (0) = 1
y" - 4y' + 4y = 64 sin 2t
y(0) = 0, y'(0) = 1
Taking Laplace transform of both sides of (1), we have
64x2
5² +4
On putting the value of y(0) and y'(0) in (2), we get
128
s²y-1-4s ÿ+4y=
s² +4
Solution:
[²y-sy)-y()]4[y-y(0)]+ 4y =
(²-4s+4)y=1+
128
s² +4
y =
or (s-2) y=1+-
1
128
1
8
16
=(5-2) (5-2) (5²+4) (-2) 3-2 (5-2)
+
8
17
8s
+
+
s-2 (S-2) s² +4
y=-8e²¹ +17e²¹ +8 cos2t
128
S +4
Ans.
+
(1)
(2)
8s
s² +4](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5a62179c-1d7b-45f7-8f2c-358abb207063%2F275fb370-a5f4-4837-b69f-677eead340e0%2Fvwaxed_processed.jpeg&w=3840&q=75)
Transcribed Image Text:EXAMPLE 12: Using Laplace transforms, find the solution of the
initial value problem
y" - 4y + 4y = 64 sin 2t
y(0) = 0, y' (0) = 1
y" - 4y' + 4y = 64 sin 2t
y(0) = 0, y'(0) = 1
Taking Laplace transform of both sides of (1), we have
64x2
5² +4
On putting the value of y(0) and y'(0) in (2), we get
128
s²y-1-4s ÿ+4y=
s² +4
Solution:
[²y-sy)-y()]4[y-y(0)]+ 4y =
(²-4s+4)y=1+
128
s² +4
y =
or (s-2) y=1+-
1
128
1
8
16
=(5-2) (5-2) (5²+4) (-2) 3-2 (5-2)
+
8
17
8s
+
+
s-2 (S-2) s² +4
y=-8e²¹ +17e²¹ +8 cos2t
128
S +4
Ans.
+
(1)
(2)
8s
s² +4
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