P(t+△)=∑Pk(t)P(△),i=1,2 k=0 于是 (+△)-P0(1)1-P0(△t () △t △t P(△t (t)+0(1)=-P0(t)+0(1) P(t+△t)-P(t) P(t)+AP-1()+0(1 得到 dPo(t) =-AP0( 0(0)=1, dP=-P()+P=1()P(0)=0.2=12,( ) ( ) ( ), 1,2, . 0 +  =   =  = − P t t P t P t i i k i i k k 于是 ( ) (1) ( ) (1). ( ) ( ) ( ) ( ) 1 ( ) 0 0 1 0 0 0 0 P t o P t o t P t P t t P t t P t t P t + = − +   = −  −  = −  +  −  ( ) ( ) (1). ( ) ( ) P t P 1 t o t P t t P t i i i i = − + +  +  −   − 得到 ( ) ( ) (0) 0, 1,2, . ( ) ( ), (0) 1, ( ) 1 0 0 0 = − + = =  = − = − P t P t P i dt dP t P t P dt dP t i i i i   