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y" + 4y = sint + u(t) sin(t − π); Hint: y(0) = 0, y'(0) = 0 since L (u₂(t)f(t-c)) = e ²³ F (³) L (U₁₂ (4) sm (t-1)) = €²³. 10 sin étts. I C=TT Here f(t) = sint 1 F (₁) = L (f (t) = L (sint) = 3 ² +1 Also for Partial fraction 1 (3²+1) (3²+4) = As+B 52 +1 + cs+D 32+4 = -TTS e 5² +1

y" + 4y = sint + u(t) sin(t − π); Hint: y(0) = 0, y'(0) = 0 since L (u₂(t)f(t-c)) = e ²³ F (³) L (U₁₂ (4) sm (t-1)) = €²³. 10 sin étts. I C=TT Here f(t) = sint 1 F (₁) = L (f (t) = L (sint) = 3 ² +1 Also for Partial fraction 1 (3²+1) (3²+4) = As+B 52 +1 + cs+D 32+4 = -TTS e 5² +1

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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Use Laplace transformation to solve. Please show and write all work neatly.

(#1 y" + 4y = sint + u„ (t) sin(t − π); Hint: L (u₂ (t) f(t-c)) = ē y(0) = 0, y'(0) = 0 écs F(s) since L (U₁ (t) sin (t-1)) = C=TT Here f(t) = sint Also for Partial flaction (3² +₁) (3² + 4) etts. I F (s) = L (f (t)) = L (simt) = 3 ² +1 As+B 5²+1 5²+1 + cstD 32² +4 -TTS 3²+1
Transcribed Image Text:(#1 y" + 4y = sint + u„ (t) sin(t − π); Hint: L (u₂ (t) f(t-c)) = ē y(0) = 0, y'(0) = 0 écs F(s) since L (U₁ (t) sin (t-1)) = C=TT Here f(t) = sint Also for Partial flaction (3² +₁) (3² + 4) etts. I F (s) = L (f (t)) = L (simt) = 3 ² +1 As+B 5²+1 5²+1 + cstD 32² +4 -TTS 3²+1
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