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Solve ODE by Laplace transform (D²−3D+2)y=e^(3t)underth…

Short answer

y(t) = (e^(3t) - e^t)/2.

Solve the differential equation (D² − 3D + 2)y = e^(3t) under the conditions y(0) = 0, y'(0) = 1, using the Laplace transform.

The equation (D² − 3D + 2)y = e^(3t) means y'' − 3y' + 2y = e^(3t), with y(0) = 0, y'(0) = 1. Let Y = L{y}.

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

Inverse transform:

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

✓ Answer checked by Claude Opus 5.5 (high effort)

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