--- title: "Solve ODE by Laplace transform y''+2y'+2y=5sinx" subject: "Mathematics III" chapter: 3 key: "semester_3/mathematics_iii:solve-ode-by-laplace-transform-y-17" source: https://www.bctengineer.com/q/semester_3/mathematics_iii/solve-ode-by-laplace-transform-y-17 site: BCT Engineering --- # Solve ODE by Laplace transform y''+2y'+2y=5sinx Mathematics III · Chapter 3: Laplace Transform - Asked in: 2079 Baishakh, 5 marks - Video: [Differential Equation - Laplace Transformation: d^2 y/dx^2 +2dy/dx+2y=5sinx ,y(0)=0 ,dy/dx=0 at x=0](https://www.youtube.com/watch?v=cHWSkVxa2Ck) — Calculus Physics Chem Accounting Tam Mai Thanh Cao - 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-y-17 - Whole chapter: https://www.bctengineer.com/md/semester_3/mathematics_iii/chapter-3 ## Question Using Laplace transform, solve the initial value problem: y'' + 2y' + 2y = 5 sin x, y(0) = y'(0) = 0. ## Short answer Using Laplace transform gives Y = 5/[(s^2+1)(s^2+2s+2)], which by partial fractions becomes (-2s+1)/(s^2+1) + (2s+3)/((s+1)^2+1). Inverting yields y(x)= sin x - 2 cos x + e^{-x}(2 cos x + sin x). ## Answer y″ + 2y′ + 2y = 5 sin x, y(0) = y′(0) = 0. Here x is the independent variable (use t in the transform). **Step 1: Transform.** With Y = L[y]: ```text (s² + 2s + 2) Y = 5/(s² + 1) Y = 5 / [ (s² + 1)(s² + 2s + 2) ] ``` **Step 2: Partial fractions.** ```text Y = (As + B)/(s² + 1) + (Cs + D)/((s + 1)² + 1) ``` Solving (compare coefficients) gives A = −2, B = 1, C = 2, D = 3: ```text Y = (−2s + 1)/(s² + 1) + (2s + 3)/((s + 1)² + 1) ``` Check of the numerator: (−2s + 1)(s² + 2s + 2) + (2s + 3)(s² + 1) = 5 ✓. **Step 3: Invert.** Write (2s + 3) = 2(s + 1) + 1: ```text y = −2 cos x + sin x + e^(−x) (2 cos x + sin x) ``` **Answer: y(x) = sin x − 2 cos x + e^(−x)(2 cos x + sin x)** **Check:** y(0) = −2 + 2 = 0 ✓. y′ = cos x + 2 sin x − e^(−x)(2 cos x + sin x) + e^(−x)(cos x − 2 sin x), so y′(0) = 1 − 2 + 1 = 0 ✓. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. (s^2 + 2s + 2)Y = A. 5/(s^2-1) B. 5s/(s^2+1) C. 5/(s^2+1) D. 5/(s^2+1) + 2 Answer: C. Zero initial values. (Key point: Transform) 2. Y = 5/[(s^2+1)(s^2+2s+2)] uses A. A/(s^2+1) + B/((s+1)^2+1) B. A/(s+1) + B/(s+2) C. As/(s^2+1) + C/(s+1) D. (As+B)/(s^2+1) + (Cs+D)/((s+1)^2+1) Answer: D. Both are irreducible quadratics. (Key point: Partial fractions form) 3. (2s+3) over (s+1)^2+1 is written as A. 2(s+1) + 1 to apply the first shift B. 2s + 3 unchanged C. 2(s-1) + 5 D. (s+1) + 2 Answer: A. Gives e^(-x)(2 cos x + sin x). (Key point: Technique) 4. y(x) = A. sin x + 2 cos x + e^(-x)(2 cos x + sin x) B. sin x - 2 cos x + e^(-x)(2 cos x + sin x) C. sin x - 2 cos x + e^(x)(2 cos x + sin x) D. sin x - 2 cos x - e^(-x)(2 cos x + sin x) Answer: B. y(0) = -2 + 2 = 0. 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