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Solve ODE by Laplace transform x''+2x'+5x=e^(−t)sint

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).

Solve the initial value problem by using Laplace transform: x'' + 2x' + 5x = e^(−t) sin t; x(0) = 0, x'(0) = 1.

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}.

(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:

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)):

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 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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