THE SCIENCE AND APPLICATIONS OF ACOUSTICS - H. H. Arnold ...

638 Appendix C. Using Laplace Transforms to Solve Differential Equationsand hencef (t) = e −at ⇔ F(s) = 1s + aC.2 Laplace Transforms of DerivativesConsider a function f (t) and its derivative df (t)/dt that are both Laplace transformable.If the function f (t) has the Laplace transform F(s), thenL df(t) = sF(s) − f (0+) (C.3)dtEquation (C.3) is readily proven by using integration by parts, i.e.,SetIt then follows thatF(s) = uv| ∞ 0F(s) =u = f (t),dv = e −st dt,∫ ∞0f (t)e −st dtdu = df(t)dtv =− 1 s e−st∫ ∞− vdu =− 1 ∣ ∣∣∣∞s f (t)e−st + 1 s0= f (0+)s+ 1 s0∫ ∞0dt∫ ∞0df(t)e −st dtdtdf(t)e −st dtdtor∫ ∞df(t)e −st dt = sF(s) − f (0+)0 dtThe term f (0+) represents the value of f (t)ast → 0 from the positive side. In asimilar manner, it can be demonstrated thatL d2 f (t)= s 2 F(s) − sf(0+) − f (0+) (C.4)dt 2C.3 Solving Differential EquationsWe discuss here a differential equation that is of greatest interest in treatment ofvibration problems:mẍ(t) + Cẋ(t) + kx(t) = f (t)(C.5)