Question

差分方程,请问图中的特解的ab是怎么求的?

Answer 1

将yt=at+b代入原方程就可以了

y(t+1) + 5 yt = 5t/2

[a(t+1)+b] + 5[at+b] = 5t/2

6at+(a+6b) = 5t/2

用待定系数法,比较两边t的系数得到

6a = 5/2

a+6b = 0

所以 a=5/12 ,b=-5/72