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Solve ODE by Laplace transform y''+4y'+3y=eᵗ

Short answer

Transform gives Y = (2s-1)/[(s-1)(s+1)(s+3)]. Partial fractions: A=1/8, B=3/4, C=-7/8. Inverse: y(t) = e^t/8 + (3/4)e^(-t) - (7/8)e^(-3t).

Solve the following initial value problem by using Laplace transform: y'' + 4y' + 3y = eᵗ, y(0) = 0, y'(0) = 2.

Transform y'' + 4y' + 3y = eᵗ with y(0) = 0, y'(0) = 2. Let Y = L{y}.

(s²Y − 0·s − 2) + 4(sY − 0) + 3Y = 1/(s − 1)
(s² + 4s + 3) Y = 2 + 1/(s − 1) = (2s − 1)/(s − 1)
Y = (2s − 1)/[(s − 1)(s + 1)(s + 3)]

Partial fractions: Y = A/(s − 1) + B/(s + 1) + C/(s + 3).

A = (2 − 1)/[(2)(4)] = 1/8
B = (−2 − 1)/[(−2)(2)] = 3/4
C = (−6 − 1)/[(−4)(−2)] = −7/8

Answer: y(t) = eᵗ/8 + (3/4) e^(−t) − (7/8) e^(−3t). Check: y(0) = 1/8 + 6/8 − 7/8 = 0 ✓; y'(0) = 1/8 − 3/4 + 21/8 = 2 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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