Skip to main content

Inverse Laplace transforms log(s/(s+1))

Short answer

a) L^{-1}{ln(s/(s+1))} = (e^{-t} - 1)/t, obtained via derivative property. b) Partial fractions give L^{-1}{1/((s-2)(s^2+1))} = (e^{2t} - cos t - 2 sin t)/5.

Find the inverse Laplace transform of: a) log(s/(s+1)) b) 1/((s − 2)(s² + 1)).

a) L⁻¹{log(s/(s + 1))}

Let F(s) = ln s − ln(s + 1), f = L⁻¹{F}. Using L{t f(t)} = −F'(s):

F'(s) = 1/s − 1/(s + 1)
L{t f(t)} = 1/(s + 1) − 1/s  → t f(t) = e^(−t) − 1
f(t) = (e^(−t) − 1)/t

b) L⁻¹{1/((s − 2)(s² + 1))}

Write 1/((s − 2)(s² + 1)) = A/(s − 2) + (Bs + C)/(s² + 1), so 1 = A(s² + 1) + (Bs + C)(s − 2).

s = 2:   1 = 5A → A = 1/5
s²:      A + B = 0 → B = −1/5
const:   A − 2C = 1 → C = −2/5
F(s) = (1/5)/(s − 2) − (1/5)(s + 2)/(s² + 1)
f(t) = (1/5)[e^(2t) − cos t − 2 sin t]

Answers: a) (e^(−t) − 1)/t; b) (e^(2t) − cos t − 2 sin t)/5.

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics III chapter 3 questions →