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Solve ODE by Laplace transform x''+4x'+4x=6e^(−t)

Short answer

Transform the ODE using initial conditions: (s+2)² X = 6/(s+1) - 2s - 16. Partial fractions give X = 6/(s+1) - 8/(s+2) - 18/(s+2)². Inverse Laplace yields x(t) = 6e^(-t) - 8e^(-2t) - 18 t e^(-2t).

Solve the following initial value problem using Laplace transform: x'' + 4x' + 4x = 6e^(−t), x(0) = −2, x'(0) = −8.

x″ + 4x′ + 4x = 6e^(−t), x(0) = −2, x′(0) = −8.

Step 1: Transform. With X = L[x]:

L[x″] = s²X − s x(0) − x′(0) = s²X + 2s + 8
L[x′] = sX − x(0)            = sX + 2
(s²X + 2s + 8) + 4(sX + 2) + 4X = 6/(s + 1)
(s + 2)² X = 6/(s + 1) − 2s − 16

Step 2: Solve for X and split.

X = 6/((s+1)(s+2)²) − (2s + 16)/(s + 2)²

6/((s+1)(s+2)²) = 6/(s+1) − 6/(s+2) − 6/(s+2)²
(2s+16)/(s+2)²  = 2/(s+2) + 12/(s+2)²
X = 6/(s+1) − 8/(s+2) − 18/(s+2)²

Step 3: Invert. L⁻¹[1/(s+2)²] = t e^(−2t).

Answer: x(t) = 6e^(−t) − 8e^(−2t) − 18t e^(−2t)

Check: x(0) = 6 − 8 = −2 ✓; x′(0) = −6 + 16 − 18 = −8 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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