例:22-1h(t)满足rh(t+5h(t+6h(t)=(t) h(0.)=h(0.)=0 ◎确定0初始值:方程两端奇异函数平衡 h(t)连续,h(0-)=h(0) (t)跃变,h(0+)≠h(0.) 方程两边积分: h"(ot+50h(t+650h()=6()dt [h(0+}-h(0.)+5[h(0+)-h(0.)+0=1 h(0+)=h(0.)=0 h(0+)=h(0.)+1=1 ⑥考虑t>0(或=0以后)的系统响应,此时激励为0P2] 齐次方程:h(t)+5h(t)+6h(t)=0 解的形式:h(t)=Ce2+C2e3tt≥0 h(t=-2Ce21-3C h(0=C1+C2=0 C1=1 h(0+)=-2C1-3C2=1 C2=- h(t)=(e2-e3)·E(t) 情况二:等号右端除t外,还有f(m)(t y (n)(t) an-ly (n-D(t).+ay((t)+ aoy(t) bmf (m)(t)+bm- f (m-D)(t)+. +bIf o(+ bof(t)8 例: 2.2-1○1 h(t) 满足 h ‘‘(t)+5 h ‘ (t)+6 h(t)=  (t) h ‘ (0-)= h (0-)=0 ○2 确定 0+初始值：方程两端奇异函数平衡 h (t)连续，h (0+) =h (0-) h ‘ (t)跃变， h ‘ (0+)≠h ‘ (0-) 方程两边积分：  + − 0 0 h ‘‘(t)dt+5  + − 0 0 h ‘ (t)dt+6  + − 0 0 h (t)=  + − 0 0  (t) dt [h‘ (0+)- h ‘ (0-)]+5[ h (0+)- h (0-)]+0=1 ∴ h (0+) =h (0-)=0 h ‘ (0+)=h‘ (0-)+1=1 ○3 考虑 t＞0(或 t=0+以后)的系统响应，此时激励为 0 [P52] 齐次方程：h ‘‘(t)+5 h ‘ (t)+6 h(t)=0 解的形式：h(t)= C1e -2t + C2e -3t t≥0 h ‘ (t)= -2C1e -2t -3C2e -3t h(0+)= C1+ C2 =0  C1 =1 h ‘ (0+)= -2C1 -3C2 =1 C2 = - 1 ∴ h(t)= (e-2t - e -3t)· (t) 情况二：等号右端除 f(t)外,还有 f (m) (t) y (n) (t)+ an-1y (n−1) (t)+…+ a1y (1) (t)+ a0y(t) = bm f (m) (t)+ bm-1 f (m−1) (t)+……+ b 1 f (1) (t)+ b0f(t)