--- title: "Inverse Laplace transforms (s+3)/(s²+6s+13)²" subject: "Mathematics III" chapter: 3 key: "semester_3/mathematics_iii:inverse-laplace-transforms-s-3-s2" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/inverse-laplace-transforms-s-3-s2 site: BCT Engineering --- # Inverse Laplace transforms (s+3)/(s²+6s+13)² Mathematics III · Chapter 3: Laplace Transform - Asked in: 2076 Chaitra, 2+3 marks - Video: [Inverse Laplace Transform Using Convolution Theorem-Laplace Transform](https://www.youtube.com/watch?v=mS130mV9XeE) — SK Maths Tutorial - 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-3-s2 - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3 ## Question Find the inverse Laplace transform of: a) (s + 3)/(s² + 6s + 13)² b) e^(−2s)/((s + 1)(s² + 2s + 2)). ## Short answer a) Complete square: (s+3)/((s+3)^2+4)^2, using shifting and L{t sin 2t} gives (1/4) t e^{-3t} sin 2t. b) Shift and second shifting: e^{-2s}/[(s+1)((s+1)^2+1)] gives e^{-(t-2)}[1 - cos(t-2)] u(t-2). ## Answer #### a) L⁻¹{(s + 3)/(s² + 6s + 13)²} Complete the square: s² + 6s + 13 = (s + 3)² + 4. So the function is (s + 3)/((s + 3)² + 4)², which is G(s + 3) with G(s) = s/(s² + 4)². L{t sin 2t} = 2(2)s/(s² + 4)² = 4s/(s² + 4)², so L⁻¹{G(s)} = (t sin 2t)/4. By the first shifting theorem (s → s + 3 gives factor e^(−3t)): ```text L⁻¹ = (1/4) t e^(−3t) sin 2t ``` #### b) L⁻¹{e^(−2s)/((s + 1)(s² + 2s + 2))} Let p = s + 1; then (s + 1)(s² + 2s + 2) = p(p² + 1) and ```text 1/(p(p² + 1)) = 1/p − p/(p² + 1) L⁻¹{1/((s+1)((s+1)²+1))} = e^(−t) (1 − cos t) ``` Second shifting theorem (e^(−as) F(s) → f(t − a) u(t − a)) with a = 2: ```text L⁻¹ = e^(−(t−2)) [1 − cos(t − 2)] u(t − 2) ``` where u is the unit step function. **Answers:** a) (1/4) t e^(−3t) sin 2t; b) e^(−(t−2))[1 − cos(t − 2)] u(t − 2). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. s^2 + 6s + 13 equals what? A. (s+3)^2 + 13 B. (s+3)^2 - 4 C. (s-3)^2 + 4 D. (s+3)^2 + 4 Answer: D. s^2 + 6s + 9 + 4. (Key point: Complete the square) 2. What is L^-1{s/(s^2+4)^2}? A. (t sin 2t)/4 B. (t sin 2t)/2 C. (sin 2t)/4 D. (t cos 2t)/4 Answer: A. L{t sin 2t} = 4s/(s^2+4)^2. (Key point: Base inverse) 3. What is L^-1{(s+3)/(s^2+6s+13)^2}? A. (1/4) t e^(3t) sin 2t B. (1/4) t e^(-3t) sin 2t C. (1/2) t e^(-3t) sin 2t D. (1/4) t e^(-3t) cos 2t Answer: B. s -> s+3 gives the factor e^(-3t). (Key point: Result for (a)) 4. What does the factor e^(-2s) do to the inverse transform? A. Multiplies it by e^(-2t) B. Advances it by 2 C. Delays it by 2 and makes it zero for t < 2 D. Differentiates it twice Answer: C. Second shifting theorem: f(t-2)u(t-2). 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