Why does the answer contain t e^(-t)?

The term 1/(s+1)^2 has a repeated pole at s = -1

Because the right side is exponential

(3/4)e^(-t) + (t/2)e^(-t) - (7/4)e^(-3t)

Solve ODE by Laplace transform (D²+4D+3)x=e^(−t)

Short answer

Apply Laplace transform, partial fractions, invert: x(t) = (7/4)e^(-t) + (t/2)e^(-t) - (3/4)e^(-3t).

Solve using Laplace transform (D² + 4D + 3)x = e^(−t), where x(0) = x'(0) = 1.

(D² + 4D + 3)x = e^(−t), x(0) = x′(0) = 1.

Step 1: Transform (X = L[x]).

(s²X − s − 1) + 4(sX − 1) + 3X = 1/(s + 1)
(s² + 4s + 3) X = 1/(s + 1) + s + 5
X = 1/[(s+1)²(s+3)] + (s + 5)/[(s+1)(s+3)]

Step 2: Partial fractions.

1/[(s+1)²(s+3)] = −1/4·1/(s+1) + 1/2·1/(s+1)² + 1/4·1/(s+3)
(s+5)/[(s+1)(s+3)] = 2/(s+1) − 1/(s+3)

Adding:

X = (7/4)/(s+1) + (1/2)/(s+1)² − (3/4)/(s+3)

Step 3: Invert.

Answer: x(t) = (7/4) e^(−t) + (t/2) e^(−t) − (3/4) e^(−3t)

Check: x(0) = 7/4 − 3/4 = 1 ✓; x′(0) = −7/4 + 1/2 + 9/4 = 1 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)