推论若[f()=F(s),则 [f"(t)]=sx[(O)]f(0) =S{S<[f(t)]f(0)}f(0 =s2<[()]-sf(0)f(0 f(t)]=sx/n1)()]-fn1(0) ="F(s)-sn-1f(0)-s22f(0)-..-n(0)(24) 特别,当初值(O)=f(0)=…m-(O)=0时,有 [f()]=sF(s),[f"(O=s2fF(s),… Ifn(DI-s"F(s) 此性质可以使我们有可能将的微分方程转 化为F()的代数方程5 推论 若L [f(t)]=F(s), 则 L [f ''(t)]=sL [f'(t)]-f '(0) =s{sL [f(t)]-f(0)}-f '(0) =s 2L [f(t)]-sf(0)-f '(0) ... L [f (n) (t)]=sL [f (n-1)(t)]-f (n-1)(0) =s nF(s)-s n-1 f(0)-s n-2 f '(0)-...-f (n-1)(0) (2.4) 特别, 当初值f(0)=f '(0)=...=f (n-1)(0)=0时, 有 L [f '(t)]=sF(s), L [f ''(t)]=s 2F(s), ..., L [f (n) (t)]=s nF(s) (2.5) 此性质可以使我们有可能将f(t)的微分方程转 化为F(s)的代数方程