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

Short answer

Transform yields (s²+4s+3)Y = s+5, so Y = 2/(s+1)−1/(s+3). Inverse Laplace gives y(t)=2e^(−t)−e^(−3t).

By using Laplace transform, solve the initial value problem y'' + 4y' + 3y = 0, given that y(0) = y'(0) = 1.

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

(s²Y − s − 1) + 4(sY − 1) + 3Y = 0
(s² + 4s + 3) Y = s + 5
Y = (s + 5)/[(s + 1)(s + 3)]

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

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

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

✓ Answer checked by Claude Opus 5.5 (high effort)

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