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Solve ODE by Laplace transform y'''+2y''−y'−2y=0

Short answer

Take Laplace transform, factor denominator as (s+2)(s−1)(s+1), use partial fractions to get Y(s)=1/(s−1) −3/(s+1) +2/(s+2). Inverse yields y(t) = eᵗ − 3e^(−t) + 2e^(−2t).

Solve the following differential equation using the Laplace transform: y''' + 2y'' − y' − 2y = 0, where y(0) = y'(0) = 0 and y''(0) = 6.

y‴ + 2y″ − y′ − 2y = 0, y(0) = y′(0) = 0, y″(0) = 6.

Step 1: Transform. With Y = L[y]:

L[y‴] = s³Y − s²y(0) − s y′(0) − y″(0) = s³Y − 6
L[y″] = s²Y,   L[y′] = sY
(s³Y − 6) + 2s²Y − sY − 2Y = 0
(s³ + 2s² − s − 2) Y = 6

Step 2: Factorise. s³ + 2s² − s − 2 = s²(s + 2) − (s + 2) = (s + 2)(s − 1)(s + 1).

Y = 6 / [(s − 1)(s + 1)(s + 2)] = A/(s−1) + B/(s+1) + C/(s+2)
A = 6/(2·3) = 1
B = 6/((−2)(1)) = −3
C = 6/((−3)(−1)) = 2

Step 3: Invert.

Answer: y(t) = eᵗ − 3e^(−t) + 2e^(−2t)

Check: y(0) = 1 − 3 + 2 = 0 ✓; y′(0) = 1 + 3 − 4 = 0 ✓; y″(0) = 1 − 3 + 8 = 6 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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