Please workout solution so the answer can be understood better 1,1. y" + y = f(t), y(0) = 0, y'(0) = 1, f(t) =0,Solution: Let Y(s) = L{y}, thenApplying the initial conditions, we obtainTherefore,S²Y (0) - 1Y (s) = L{f(t)}.The forcing function f(t) can be written as f(t) = 1 - u3 (t). Therefore,L{f(t)} = L{1U3n(t)} = L{1} − L{U3n(t)}or using partial fractions, we haveY(s) =Thus,(b)0.80.6s²Y (s) sy(0) - y'(0) +Y(s) = L{f(t)}.0.40.2Y(s)18² +1 S+120≤t<3π,3π ≤t<∞0,1s² +1116+1==S1-e-3nss(s²+1)Ss²+1-3TSS+-3πSSse-3nss² + 1y(t) = sin(t) + 1 = cos(t) - u3n(t) + u3n (t) cos(t - 3π).1-p-3nsSMA1618

Answered: Image /qna-images/answer/b57a9681-7bc4-4372-935e-27cca16d0b0c.jpg

1. y"+y = f(t), y(0) = 0, y'(0) = 1, f(t) = Solution: Let Y(s) = L{y}, then Applying the initial conditions, we obtain Therefore, S²Y (0) 1Y (s) = L{f(t)}. The forcing function f(t) can be written as f(t)=1u3r(t). Therefore, L{f(t)} = {1- u3n (t)} = L{1} - L{U3n(t)} or using partial fractions, we have Thus, (b) 0.8 0.6 0,4 s²Y(s) - sy(0) - y'(0) +Y(s) = L{f(t)}. 0.2 0 Y(s) = Y(s) 1 s²+1 10 1, 0≤t<3π, 3π

1. y"+y = f(t), y(0) = 0, y'(0) = 1, f(t) = Solution: Let Y(s) = L{y}, then Applying the initial conditions, we obtain Therefore, S²Y (0) 1Y (s) = L{f(t)}. The forcing function f(t) can be written as f(t)=1u3r(t). Therefore, L{f(t)} = {1- u3n (t)} = L{1} - L{U3n(t)} or using partial fractions, we have Thus, (b) 0.8 0.6 0,4 s²Y(s) - sy(0) - y'(0) +Y(s) = L{f(t)}. 0.2 0 Y(s) = Y(s) 1 s²+1 10 1, 0≤t<3π, 3π

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.

Chapter14: Discrete Dynamical Systems

Section14.3: Determining Stability

Problem 13E: Repeat the instruction of Exercise 11 for the function. f(x)=x3+x For part d, use i. a1=0.1 ii...

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