Please workout solution so the answer can be understood better 3. y" + 4y = sin(t) − u2 (t) sin(t - 2π), y(0) = 0, y'(0) = 0Solution: Let Y(s) = L{y}, then applying the Laplace transform to the equation, we obtains²Y (s)-sy(0) - y'(0) + 4Y(s) =Therefore,or by partial fractionsDifferential EquationsTherefore,y(t)1-e-2πsY(s) = (s² + 4) (s² + 1)=1Y(s) = (1-6²) [21-2² +₁]Section 6.411[sin(tsin(t) — sin(2t) — u27 (t) sin(t − 2π) + -u27 (t) sin(2(t – 2π))= (1 − u2# (t))[2 sin(t) — sin(2t)]101-e-2πss²+10.40.30.20.1A-0.3-0.4-Homework Solutions10

Answered: Image /qna-images/answer/f3a25d7e-91cb-40ed-82cf-309f2020728b.jpg

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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