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Solve ODE by Laplace transform y''−3y'+2y=4x+e^(3x)

Short answer

y(x) = 3 + 2x - e^x/2 - 2 e^(2x) + e^(3x)/2.

Solve the initial value problem by the Laplace transform method: y'' − 3y' + 2y = 4x + e^(3x), y(0) = 1, y'(0) = −1.

The independent variable is x; use t in the transform. Solve y'' − 3y' + 2y = 4x + e^(3x), y(0) = 1, y'(0) = −1. Let Y = L{y}.

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

Partial fractions of each part:

(s − 4)/[(s − 1)(s − 2)]      = 3/(s − 1) − 2/(s − 2)
4/[s²(s − 1)(s − 2)]          = 3/s + 2/s² − 4/(s − 1) + 1/(s − 2)
1/[(s − 1)(s − 2)(s − 3)]     = (1/2)/(s − 1) − 1/(s − 2) + (1/2)/(s − 3)

Adding:

Y = 3/s + 2/s² − (1/2)/(s − 1) − 2/(s − 2) + (1/2)/(s − 3)

Inverse transform:

Answer: y(x) = 3 + 2x − eˣ/2 − 2e^(2x) + e^(3x)/2. Check: y(0) = 3 − 1/2 − 2 + 1/2 = 1 ✓; y'(0) = 2 − 1/2 − 4 + 3/2 = −1 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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