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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

Advanced Engineering Mathematics
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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1, 1. y" + y = f(t), y(0) = 0, y'(0) = 1, f(t) = 0, 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) = 1 - u3 (t). Therefore, L{f(t)} = L{1U3n(t)} = L{1} − L{U3n(t)} or using partial fractions, we have Y(s) = Thus, (b) 0.8 0.6 s²Y (s) sy(0) - y'(0) +Y(s) = L{f(t)}. 0.4 0.2 Y(s) 1 8² +1 S + 12 0≤t<3π, 3π ≤t<∞0, 1 s² +1 1 16 + 1 == S 1-e-3ns s(s²+1) S s²+1 -3TS S + -3πS S se-3ns s² + 1 y(t) = sin(t) + 1 = cos(t) - u3n(t) + u3n (t) cos(t - 3π). 1-p-3ns S MA 16 18
Transcribed Image Text:1, 1. y" + y = f(t), y(0) = 0, y'(0) = 1, f(t) = 0, 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) = 1 - u3 (t). Therefore, L{f(t)} = L{1U3n(t)} = L{1} − L{U3n(t)} or using partial fractions, we have Y(s) = Thus, (b) 0.8 0.6 s²Y (s) sy(0) - y'(0) +Y(s) = L{f(t)}. 0.4 0.2 Y(s) 1 8² +1 S + 12 0≤t<3π, 3π ≤t<∞0, 1 s² +1 1 16 + 1 == S 1-e-3ns s(s²+1) S s²+1 -3TS S + -3πS S se-3ns s² + 1 y(t) = sin(t) + 1 = cos(t) - u3n(t) + u3n (t) cos(t - 3π). 1-p-3ns S MA 16 18
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