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Solve ODE by Laplace transform y''+4y'+3y=0

Short answer

Laplace transform leads to Y = (3s+13)/[(s+1)(s+3)]. Partial fractions give 5/(s+1) - 2/(s+3); inverting yields y(t) = 5e^(-t) - 2e^(-3t).

Solve the initial value problem y'' + 4y' + 3y = 0, y(0) = 3, y'(0) = 1 by using Laplace transform.

Transform y'' + 4y' + 3y = 0 with y(0) = 3, y'(0) = 1. Let Y = L{y}.

L{y''} = s²Y − 3s − 1
L{y'}  = sY − 3

(s²Y − 3s − 1) + 4(sY − 3) + 3Y = 0
(s² + 4s + 3) Y = 3s + 13
Y = (3s + 13)/[(s + 1)(s + 3)]

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

A = (3(−1) + 13)/(−1 + 3) = 10/2 = 5
B = (3(−3) + 13)/(−3 + 1) = 4/(−2) = −2
Y = 5/(s + 1) − 2/(s + 3)

Answer: y(t) = 5e^(−t) − 2e^(−3t). Check: y(0) = 3 ✓; y'(0) = −5 + 6 = 1 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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