--- title: "Inverse Laplace transforms (s+2)³/s⁴" subject: "Mathematics III" chapter: 3 key: "semester_3/mathematics_iii:inverse-laplace-transforms-s-2-3" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/inverse-laplace-transforms-s-2-3 site: BCT Engineering --- # Inverse Laplace transforms (s+2)³/s⁴ Mathematics III · Chapter 3: Laplace Transform - Asked in: 2070 Asar, 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-s-2-3 - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3 ## Question Find the inverse Laplace transform of (i) (s + 2)³/s⁴ (ii) 1/(s²(s² + a²)). ## Short answer Expand (s+2)^3/s^4 = 1/s +6/s^2 +12/s^3 +8/s^4, inverse is 1 +6t +6t^2 +(4/3)t^3. For 1/(s^2(s^2+a^2)), partial fractions (1/a^2)(1/s^2 -1/(s^2+a^2)) yield inverse (at - sin at)/a^3. ## Answer #### (i) L⁻¹[(s + 2)³/s⁴] Expand: (s + 2)³ = s³ + 6s² + 12s + 8. ```text (s+2)³/s⁴ = 1/s + 6/s² + 12/s³ + 8/s⁴ ``` Use L⁻¹[1/sⁿ⁺¹] = tⁿ/n!: ```text = 1 + 6t + 12·t²/2 + 8·t³/6 ``` **Answer: 1 + 6t + 6t² + (4/3)t³** #### (ii) L⁻¹[1/(s²(s² + a²))] ```text 1/(s²(s²+a²)) = (1/a²) [ 1/s² − 1/(s² + a²) ] ``` **Answer: (1/a²)[ t − (sin at)/a ] = (at − sin at)/a³** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. (s+2)^3/s^4 = A. 1/s + 3/s^2 + 3/s^3 + 8/s^4 B. 1/s + 6/s^2 + 12/s^3 + 8/s^3 C. 1/s + 6/s^2 + 12/s^3 + 8/s^4 D. s + 6 + 12/s + 8/s^2 Answer: C. (s+2)^3 = s^3 + 6s^2 + 12s + 8. (Key point: Expansion) 2. L^-1[1/s^(n+1)] = A. t^n B. n! t^n C. t^(n+1)/n! D. t^n/n! Answer: D. Standard. (Key point: Inverse rule) 3. L^-1 = A. 1 + 6t + 6t^2 + (4/3)t^3 B. 1 + 6t + 12t^2 + 8t^3 C. 1 + 6t + 6t^2 + 8t^3 D. 1 + 3t + 3t^2 + t^3 Answer: A. 12 t^2/2 = 6t^2, 8 t^3/6 = 4t^3/3. (Key point: Result (i)) 4. L^-1[1/(s^2(s^2+a^2))] = A. (at + sin at)/a^3 B. (at - sin at)/a^3 C. (t - sin at)/a^2 D. (at - sin at)/a^2 Answer: B. (1/a^2)[t - sin(at)/a]. 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