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- 2070 Asar · 5 marks
Inverse Laplace transforms (s+2)³/s⁴
Short answer
Expand (s+2)^3/s^4 = 1/s +6/s^2 +12/s^3 +8/s^4, inverse is 1 +6t +6t^2 +(4/3)t^3. For 1/(s^2(s^2+a^2)), partial fractions (1/a^2)(1/s^2 -1/(s^2+a^2)) yield inverse (at - sin at)/a^3.
Find the inverse Laplace transform of (i) (s + 2)³/s⁴ (ii) 1/(s²(s² + a²)).
(i) L⁻¹[(s + 2)³/s⁴]
Expand: (s + 2)³ = s³ + 6s² + 12s + 8.
(s+2)³/s⁴ = 1/s + 6/s² + 12/s³ + 8/s⁴
Use L⁻¹[1/sⁿ⁺¹] = tⁿ/n!:
= 1 + 6t + 12·t²/2 + 8·t³/6
Answer: 1 + 6t + 6t² + (4/3)t³
(ii) L⁻¹[1/(s²(s² + a²))]
1/(s²(s²+a²)) = (1/a²) [ 1/s² − 1/(s² + a²) ]
Answer: (1/a²)[ t − (sin at)/a ] = (at − sin at)/a³
✓ Answer checked by Claude Opus 5.5 (high effort)
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