--- title: "Inverse Laplace transforms (2s+3)/(s²+5s−6)" subject: "Mathematics III" chapter: 3 key: "semester_3/mathematics_iii:inverse-laplace-transforms-2s-3-s2" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/inverse-laplace-transforms-2s-3-s2 site: BCT Engineering --- # Inverse Laplace transforms (2s+3)/(s²+5s−6) Mathematics III · Chapter 3: Laplace Transform - Asked in: 2070 Chaitra, 2.5+2.5 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-2s-3-s2 - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3 ## Question Find the inverse Laplace transform of (a) (2s + 3)/(s² + 5s − 6) (b) s³/(s⁴ − a⁴). ## Short answer a) Partial fractions: (5/7)e^t + (9/7)e^{-6t}. b) Split as ½[s/(s^2-a^2)+s/(s^2+a^2)] = ½(cosh at + cos at). ## Answer #### (a) L⁻¹[(2s + 3)/(s² + 5s − 6)] s² + 5s − 6 = (s + 6)(s − 1). ```text (2s + 3)/((s−1)(s+6)) = A/(s−1) + B/(s+6) A = (2·1 + 3)/(1 + 6) = 5/7 B = (−12 + 3)/(−6 − 1) = 9/7 ``` **Answer: (5/7) eᵗ + (9/7) e^(−6t)** #### (b) L⁻¹[s³/(s⁴ − a⁴)] s⁴ − a⁴ = (s² − a²)(s² + a²). Then ```text s³/((s²−a²)(s²+a²)) = ½ [ s/(s² − a²) + s/(s² + a²) ] ``` (Check: s(s²+a²) + s(s²−a²) = 2s³ ✓.) **Answer: ½ (cosh at + cos at)** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. s^2+5s-6 = ? A. (s-6)(s+1) B. (s+6)(s-1) C. (s+3)(s-2) D. (s+6)(s+1) Answer: B. Product -6 sum 5. (Key point: Factor) 2. A for 1/(s-1) is? A. 5/6 B. 7/5 C. 5/7 D. 1/7 Answer: C. (2+3)/(1+6). (Key point: A) 3. What is the inverse? A. (5/7)e^(-t) + (9/7)e^(6t) B. (9/7)e^t + (5/7)e^(-6t) C. (5/7)e^t - (9/7)e^(-6t) D. (5/7)e^t + (9/7)e^(-6t) Answer: D. A and B. (Key point: Result for (a)) 4. What is L^-1{s^3/(s^4-a^4)}? A. (1/2)(cosh at + cos at) B. (1/2)(sinh at + sin at) C. (1/2)(cosh at - cos at) D. cosh at + cos at Answer: A. (1/2)[s/(s^2-a^2) + s/(s^2+a^2)]. 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