--- title: "Inverse Laplace transforms s²/(s−1)³" subject: "Mathematics III" chapter: 3 key: "semester_3/mathematics_iii:inverse-laplace-transforms-s2-s-1" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/inverse-laplace-transforms-s2-s-1 site: BCT Engineering --- # Inverse Laplace transforms s²/(s−1)³ Mathematics III · Chapter 3: Laplace Transform - Asked in: 2081 Baishakh, 2+3 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/inverse-laplace-transforms-s2-s-1 - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3 ## Question Find the inverse Laplace transform of a) s²/(s − 1)³ b) s²/(s² + 4)². ## Short answer a) e^t (1 + 2t + t^2/2); b) (1/4) sin 2t + (1/2) t cos 2t. ## Answer #### a) L⁻¹{s²/(s − 1)³} Write s² = (s − 1)² + 2(s − 1) + 1: ```text s²/(s − 1)³ = 1/(s − 1) + 2/(s − 1)² + 1/(s − 1)³ L⁻¹ = eᵗ + 2t eᵗ + (t²/2) eᵗ = eᵗ (1 + 2t + t²/2) ``` #### b) L⁻¹{s²/(s² + 4)²} Since L{t sin 2t} = 4s/(s² + 4)², we have L⁻¹{s/(s² + 4)²} = (t sin 2t)/4 =: g(t), with g(0) = 0. Then s · G(s) = L{g'(t)} (because g(0) = 0): ```text g'(t) = (sin 2t)/4 + (t cos 2t)/2 ``` So L⁻¹{s²/(s² + 4)²} = (1/4) sin 2t + (1/2) t cos 2t. **Answers:** a) eᵗ (1 + 2t + t²/2); b) (sin 2t)/4 + (t cos 2t)/2. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Write s^2 in powers of (s-1). A. (s-1)^2 + (s-1) + 1 B. (s-1)^2 + 2(s-1) + 1 C. (s-1)^2 - 2(s-1) + 1 D. (s-1)^2 + 2(s-1) - 1 Answer: B. Expanding gives s^2 - 2s + 1 + 2s - 2 + 1 = s^2. (Key point: Rewrite s^2) 2. What is L^-1{s^2/(s-1)^3}? A. e^t (1 + 2t + t^2) B. e^t (1 + t + t^2/2) C. e^t (1 + 2t + t^2/2) D. e^(-t) (1 + 2t + t^2/2) Answer: C. Terms: e^t, 2t e^t, and (t^2/2) e^t. (Key point: Part (a)) 3. Which known transform is used to invert s^2/(s^2+4)^2? A. L{t sin 2t} = 2s/(s^2+4)^2, so L^-1{s/(s^2+4)^2} = t sin 2t / 2 B. L{t cos 2t} = 4s/(s^2+4)^2 C. L{sin 2t} = s/(s^2+4)^2 D. L{t sin 2t} = 4s/(s^2+4)^2, so L^-1{s/(s^2+4)^2} = t sin 2t / 4 Answer: D. Using L{t sin kt} = 2ks/(s^2+k^2)^2 with k = 2. (Key point: Starting point for part (b)) 4. What is L^-1{s^2/(s^2+4)^2}? A. (sin 2t)/4 + (t cos 2t)/2 B. (sin 2t)/4 - (t cos 2t)/2 C. (t sin 2t)/4 D. (cos 2t)/4 + (t sin 2t)/2 Answer: A. s*G(s) = L{g'} since g(0) = 0, and g' of (t sin 2t)/4 is (sin 2t)/4 + (t cos 2t)/2. 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