特解:为响应的稳态分量或强制分量)p=s 所以过阻尼情况=bb+1=ke+ke+is 定常数/+k+=0解得 S1-S2 LS,K,+s,k2=0特解：为响应的稳态分量(或强制分量)iLp=iS 所以过阻尼情况 1 2 1 2 s t s t L Lh Lp S i i i k e k e i = + = + + 定常数 1 2 1 1 2 2 0 0 S k k i s k s k  + + =   + = 解得 2 1 1 2 1 2 2 1 S S s k i s s s k i s s  =   −   =  − L i 1 0 t