解:(b)y(-1)=0的情况 x, (n=d y(n-ay(n n(0)=ay(-1)+(0)=1 (1)=ay1(0)+(1)=a y(2)=a()+6(2)=a2 y(n)=ay1(n-1)+(n)=a 所以:y(n)=a"lv(n)解：（b）y(−1) = 0的情况 ( ) ( ) 令 x1 n = n y1 (0) = ay1 (−1)+ (0) =1 y1 (1) = ay1 (0)+ (1) = a 2 1 1 y (2) = ay (1) + (2) = a …. n y1 (n) = ay1 (n −1) + (n) = a 所以： ( ) ( ) y1 n a u n n =  y(n) − ay(n −1) = x(n)