例4:方程为:r(t)+3r(t)+2r(t)=e(t)+2(t) 若激励为:e(t)=t2求其特解r). 查表23-1得对应的特征解为:r(t)=A,t+At+A) (),(),r()e(t),e(t)代入原微分方程得: 24+324,t+4)+2(4t2+4t+A)=2t+22 241+(64,+2A)t+(24,+34t+2A)=2t+22 等式两边同次幂系数相等:2A,=2 A=2 6A+2A=2 →A=-2 24,+3A+2A=0 A,=1 rr(t)=t2-2t+2 t≥0例4:方程为: ( ) 3 ( ) 2 ( ) ( ) 2 ( ) " ' ' r t r t r t e t e t 若激励为: 2 e(t) t 求其特解 rf (t). 查表2-3-1得对应的特征解为: 1 0 2 2 rf (t) A t At A ( ), ( ), ( ) " ' r t r t r t f f f ( ), ( ) ' e t e t 代入原微分方程得: 2 1 0 2 2 2 1 2 2A 3(2At A)2(At At A ) 2t 2t 2 2 1 2 1 0 2 2 2At (6A 2A)t (2A 3At 2A ) 2t 2t 等式两边同次幂系数相等: 1 2 2 2 3 2 0 6 2 2 2 2 2 1 0 2 1 0 2 1 2 A A A A A A A A A ( ) 2 2 2 rf t t t t 0