Vector

◗ Theorem 12.18 (Division by t) Let both f (t) and f(t)thave Laplace transformsand let F(s) denote the transform of f (t).If limt→0+f(t)texists, thenL(f(t)t=s∞F(σ) dσ. Proof BecauseF(σ) =0∞f(t) e-σtdt,we integrateF(σ) from s to∞ and

s∞F (σ)dσ =o∞0∞f (t) e-σtdtdσ.

We reverse the order of integration in the double integral of this equation to Obtain

s∞F (σ) dσ=0∞s∞f(t)e-σtdσdt = 0∞-f(t)te-σt ? dt =0∞f(t)te-stdt=Lf(t)t. Tapos na.

Example 12.21. Show that

.

Solution. If we let , then to obtain the desired result:

Example 12.22. Show that Solution.

. Hence we can differentiate

.

We let

, then

integrate

. Because

, we can

to obtain the desired result:

Laplace Transform III: Second order equations; completing the square. Rules: L is linear: L[af(t) + bg(t)] = aF(s) + bG(s)F(s) essentially determines f(t)s-shift: L[e^{at}f(t)] = F(s-a)t-shift: L[f_a(t)] = e^{-as} F(s)s-derivative: L[tf(t)] = - F'(s)t-derivative: L[f'(t)] = s F(s) - f(0+) L[f"(t)] = s^2 F(s) - s f(0+) - f'(0+) (ignoring singularities at t = 0 ) Computations : L[1] 1/s = L[e^{as}] L[cos(omega t)] L[sin(omega t)] L[delta(t-a)] L[t^n]

= = = = =

1/(s-a) s/(s^2+omega^2) omega/(s^2+omega^2)e^{-as}

n!/s^{n+1} , n = 1, 2, 3, ...