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

Short answer

Set p = s+1; the transformed equation yields Y = 1/(p^2+4) + 1/[(p^2+1)(p^2+4)]. Simplifying and applying the first shift theorem gives y(t) = (1/3)e^(-t)(sin t + sin 2t).

Apply Laplace transform to solve the differential equation y'' + 2y' + 5y = e^(−t) sin t, x(0) = 0, x'(0) = 1.

Transform y'' + 2y' + 5y = e^(−t) sin t with y(0) = 0, y'(0) = 1. (The question writes the unknown as both y and x; it is the same function of t, so y is used here.) Let Y = L{y}.

(s²Y − 1) + 2sY + 5Y = L{e^(−t) sin t} = 1/((s + 1)² + 1)
(s² + 2s + 5) Y = 1 + 1/((s + 1)² + 1)

Put p = s + 1, so s² + 2s + 5 = p² + 4:

Y = 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)):

y(t) = e^(−t) [ (1/3) sin t + (2/3)(sin 2t)/2 ]

Answer: y(t) = (1/3) e^(−t) (sin t + sin 2t). Check: y(0) = 0; y'(0) = (1/3)(1 + 2) = 1 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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