Solve ODE by Laplace transform y''+y'−2y=x

The independent variable is x here; write it as t for the Laplace transform. Solve y'' + y' − 2y = t with y(0) = 1, y'(0) = 0. Let Y = L{y}.

(s²Y − s) + (sY − 1) − 2Y = 1/s²
(s² + s − 2) Y = s + 1 + 1/s² = (s³ + s² + 1)/s²
Y = (s³ + s² + 1)/[s² (s − 1)(s + 2)]

Partial fractions: Y = A/s + B/s² + C/(s − 1) + D/(s + 2).

B = 1/[(−1)(2)] = −1/2                    (s = 0 in the other factors)
C = (1 + 1 + 1)/[1 · 3] = 1               (s = 1)
D = (−8 + 4 + 1)/[4 · (−3)] = 1/4         (s = −2)
As s → ∞, Y ≈ 1/s, so A + C + D = 1 → A = 1 − 1 − 1/4 = −1/4
Y = −1/(4s) − 1/(2s²) + 1/(s − 1) + 1/(4(s + 2))

Inverse transform:

y = −1/4 − x/2 + eˣ + (1/4) e^(−2x)

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