3. y" + 4y = sin(t) − u2 (t) sin(t - 2π), y(0) = 0, y'(0) = 0 Solution: Let Y(s) = L{y}, then applying the Laplace transform to the equation, we obtain Therefore, or by partial fractions Differential Equations Therefore, (b) y(t) = = s²Y (s)-sy(0) - y'(0) + 4Y(s) = Y(s) = 1-e-2ns (s² + 4) (s² + 1) Y(s) = (1-6²) [21-2² +₁] e Section 6.4 (1 − u2x (t))[2 sin(t) — sin(2t)] 10 1 1 sin(t) sin(t) — sin(2t) — u27 (t) sin(t − 2π) +-u2 (t) sin(2(t − 2π)) - 12 1-e-2ns s² +1 0.4 0.3 0.2 0.1- A 0 -0.1 -0.2 -0.3 -0.4- Homework Solutions 10 12
3. y" + 4y = sin(t) − u2 (t) sin(t - 2π), y(0) = 0, y'(0) = 0 Solution: Let Y(s) = L{y}, then applying the Laplace transform to the equation, we obtain Therefore, or by partial fractions Differential Equations Therefore, (b) y(t) = = s²Y (s)-sy(0) - y'(0) + 4Y(s) = Y(s) = 1-e-2ns (s² + 4) (s² + 1) Y(s) = (1-6²) [21-2² +₁] e Section 6.4 (1 − u2x (t))[2 sin(t) — sin(2t)] 10 1 1 sin(t) sin(t) — sin(2t) — u27 (t) sin(t − 2π) +-u2 (t) sin(2(t − 2π)) - 12 1-e-2ns s² +1 0.4 0.3 0.2 0.1- A 0 -0.1 -0.2 -0.3 -0.4- Homework Solutions 10 12
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter6: Applications Of The Derivative
Section6.3: Implicit Differentiation
Problem 45E
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![3. y" + 4y = sin(t) − u2 (t) sin(t - 2π), y(0) = 0, y'(0) = 0
Solution: Let Y(s) = L{y}, then applying the Laplace transform to the equation, we obtain
s²Y (s)-sy(0) - y'(0) + 4Y(s) =
Therefore,
or by partial fractions
Differential Equations
Therefore,
y(t)
1-e-2πs
Y(s) = (s² + 4) (s² + 1)
=
1
Y(s) = (1-6²) [21-2² +₁]
Section 6.4
1
1
[sin(t
sin(t) — sin(2t) — u27 (t) sin(t − 2π) + -u27 (t) sin(2(t – 2π))
= (1 − u2# (t))[2 sin(t) — sin(2t)]
10
1-e-2πs
s²+1
0.4
0.3
0.2
0.1
A
-0.3
-0.4-
Homework Solutions
10](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F8ce6f6e3-a53e-4898-b247-4366f6e6c1b7%2Ff3a25d7e-91cb-40ed-82cf-309f2020728b%2F4rfm0ep_processed.jpeg&w=3840&q=75)
Transcribed Image Text:3. y" + 4y = sin(t) − u2 (t) sin(t - 2π), y(0) = 0, y'(0) = 0
Solution: Let Y(s) = L{y}, then applying the Laplace transform to the equation, we obtain
s²Y (s)-sy(0) - y'(0) + 4Y(s) =
Therefore,
or by partial fractions
Differential Equations
Therefore,
y(t)
1-e-2πs
Y(s) = (s² + 4) (s² + 1)
=
1
Y(s) = (1-6²) [21-2² +₁]
Section 6.4
1
1
[sin(t
sin(t) — sin(2t) — u27 (t) sin(t − 2π) + -u27 (t) sin(2(t – 2π))
= (1 − u2# (t))[2 sin(t) — sin(2t)]
10
1-e-2πs
s²+1
0.4
0.3
0.2
0.1
A
-0.3
-0.4-
Homework Solutions
10
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