Skip to main content

Laplace transforms of functions e^(2t)cos²2t

Short answer

Existence: f must be piecewise continuous on finite intervals and of exponential order. i) cos²2t = (1+cos4t)/2, shift s→s−2: ½[1/(s−2) + (s−2)/(( s−2)²+16)]. ii) L{(cos2t−cos3t)/t} = ∫ₛ^∞ (u/(u²+4)−u/(u²+9)) du = ½ ln((s²+9)/(s²+4)).

State the conditions for existence of Laplace transform. Obtain the Laplace transform of: (i) e^(2t) cos² 2t (ii) (cos 2t − cos 3t)/t.

Conditions for existence

The Laplace transform L[f(t)] = ∫₀^∞ e^(−st) f(t) dt exists if

  1. f(t) is piecewise continuous on every finite interval [0, A], and
  2. f(t) is of exponential order, that is |f(t)| ≤ M e^(kt) for all t > T, for some constants M, k, T.

These conditions are sufficient (not necessary); then L[f] exists for s > k.

(i) L[e^(2t) cos² 2t]

cos² 2t = (1 + cos 4t)/2, so

L[cos² 2t] = ½ [1/s + s/(s² + 16)]

By the first shifting theorem, replace s by (s − 2):

L[e^(2t) cos² 2t] = ½ [ 1/(s − 2) + (s − 2)/((s − 2)² + 16) ]

Answer: ½ [1/(s − 2) + (s − 2)/((s − 2)² + 16)]

(ii) L[(cos 2t − cos 3t)/t]

Use L[f(t)/t] = ∫ₛ^∞ F(u) du with F(u) = u/(u² + 4) − u/(u² + 9).

∫ₛ^∞ [u/(u²+4) − u/(u²+9)] du = ½ [ ln((u²+4)/(u²+9)) ]ₛ^∞
                              = ½ [ 0 − ln((s²+4)/(s²+9)) ]

Answer: ½ ln((s² + 9)/(s² + 4))

✓ Answer checked by Claude Opus 5.5 (high effort)

Related questions

All Mathematics III chapter 3 questions →