--- title: "Solve ODE by Laplace transform x''+2x'+5x=e^(−t)sint" subject: "Mathematics III" chapter: 3 key: "semester_3/mathematics_iii:solve-ode-by-laplace-transform-x" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/solve-ode-by-laplace-transform-x site: BCT Engineering --- # Solve ODE by Laplace transform x''+2x'+5x=e^(−t)sint Mathematics III · Chapter 3: Laplace Transform - Asked in: 2072 Kartik, 5 marks; 2069 Asar, 5 marks (asked 2 times) - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/mathematics_iii/solve-ode-by-laplace-transform-x - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3 ## Question Solve the initial value problem by using Laplace transform: x'' + 2x' + 5x = e^(−t) sin t; x(0) = 0, x'(0) = 1. ## Short answer With p = s+1, transform gives X = (1/3)/(p^2+1) + (2/3)/(p^2+4). Inverse transform using the first shift theorem: x(t) = (1/3)e^(-t)(sin t + sin 2t). ## Answer Transform x'' + 2x' + 5x = e^(−t) sin t with x(0) = 0, x'(0) = 1. The unknown is x(t); write X = X(s) = L{x}. ```text (s²X − 1) + 2sX + 5X = L{e^(−t) sin t} = 1/((s + 1)² + 1) (s² + 2s + 5) X = 1 + 1/((s + 1)² + 1) ``` Put p = s + 1, so s² + 2s + 5 = p² + 4: ```text X = 1/(p² + 4) + 1/[(p² + 1)(p² + 4)] = 1/(p² + 4) + (1/3)[1/(p² + 1) − 1/(p² + 4)] = (1/3) · 1/(p² + 1) + (2/3) · 1/(p² + 4) ``` Inverse transform (first shifting theorem, s → p = s + 1 gives factor e^(−t)): ```text x(t) = e^(−t) [ (1/3) sin t + (2/3)(sin 2t)/2 ] ``` **Answer: x(t) = (1/3) e^(−t) (sin t + sin 2t).** Check: x(0) = 0; x'(0) = (1/3)(1 + 2) = 1 ✓. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With x(0) = 0 and x'(0) = 1, what is L{x''}? A. s^2 X - s B. s^2 X - s - 1 C. s^2 X + 1 D. s^2 X - 1 Answer: D. L{x''} = s^2 X - s x(0) - x'(0) = s^2 X - 1. (Key point: Transform of x'') 2. What is L{e^(-t) sin t}? A. 1/((s+1)^2 + 1) B. 1/(s^2 + 1) C. 1/((s-1)^2 + 1) D. (s+1)/((s+1)^2 + 1) Answer: A. First shifting replaces s by s+1 in 1/(s^2+1). (Key point: Transform of the forcing term) 3. s^2 + 2s + 5 equals what? A. (s+1)^2 + 5 B. (s+1)^2 + 4 C. (s-1)^2 + 4 D. (s+1)^2 - 4 Answer: B. With p = s+1 the denominator becomes p^2 + 4. (Key point: Completing the square) 4. (1/3)/(p^2+1) + (2/3)/(p^2+4), p = s+1, inverts to what? A. e^(-t) [ (1/3) sin t + (2/3) sin 2t ] B. e^(t) [ (1/3) sin t + (1/3) sin 2t ] C. e^(-t) [ (1/3) sin t + (1/3) sin 2t ] D. e^(-t) [ (1/3) cos t + (1/3) cos 2t ] Answer: C. L^-1{2/(3(p^2+4))} = (1/3) sin 2t, and the shift contributes e^(-t). 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