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Solve ODE by Laplace transform y''−10y'+9y=5t

Short answer

Apply Laplace transform and partial fractions; solution: y(t) = 50/81 + 5t/9 - 2e^t + (31/81) e^{9t}.

Solve the initial value problem by applying the Laplace transform: y'' − 10y' + 9y = 5t, y(0) = −1, y'(0) = 2.

Transform y'' − 10y' + 9y = 5t with y(0) = −1, y'(0) = 2. Let Y = L{y}.

(s²Y + s − 2) − 10(sY + 1) + 9Y = 5/s²
(s² − 10s + 9) Y = −s + 12 + 5/s²
Y = (12 − s)/[(s − 1)(s − 9)] + 5/[s²(s − 1)(s − 9)]

Partial fractions:

(12 − s)/[(s − 1)(s − 9)] = (−11/8)/(s − 1) + (3/8)/(s − 9)
5/[s²(s − 1)(s − 9)]      = (50/81)/s + (5/9)/s² − (5/8)/(s − 1) + (5/648)/(s − 9)

Y = (50/81)/s + (5/9)/s² − 2/(s − 1) + (31/81)/(s − 9)

Inverse transform:

Answer: y(t) = 50/81 + 5t/9 − 2eᵗ + (31/81) e^(9t). Check: y(0) = 50/81 − 2 + 31/81 = −1 ✓; y'(0) = 5/9 − 2 + 31/9 = 2 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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