Skip to main content

Inverse Laplace transforms

Short answer

1/(s^2-5s+6) = 1/(( s-3)(s-2)) = 1/(s-3)-1/(s-2), inverse is e^(3t)-e^(2t). For (s+2)/(s^2+4s+5)^2, complete square (s+2)^2+1, shift and use L^-1[s/(s^2+1)^2] = (t sin t)/2, yielding ½ t e^(-2t) sin t.

Find the inverse Laplace transform of the following: a) 1/(s² − 5s + 6) b) (s + 2)/(s² + 4s + 5)².

a) L⁻¹[1/(s² − 5s + 6)]

s² − 5s + 6 = (s − 2)(s − 3), and 1/((s−2)(s−3)) = 1/(s − 3) − 1/(s − 2).

Answer: e^(3t) − e^(2t)

b) L⁻¹[(s + 2)/(s² + 4s + 5)²]

Complete the square: s² + 4s + 5 = (s + 2)² + 1. So F(s) = G(s + 2) with G(s) = s/(s² + 1)².

Now L[t sin t] = 2s/(s² + 1)², hence L⁻¹[s/(s²+1)²] = (t sin t)/2. (Or by convolution of cos t and sin t.)

By the first shifting theorem, L⁻¹[G(s + 2)] = e^(−2t) g(t):

Answer: ½ t e^(−2t) sin t

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics III chapter 3 questions →