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y'' 2y' 2y f',y'(0 ) 1,y(0 ) 0,f (t)解:1.求yx(t)

2 2 2 0, 1,2 1 j

yx(t) e t(Cx1cost Cx2sint)

y'x(t) e t(Cx2cost Cx1sint) e t(Cx1cost Cx2sint)代入初始状态：yx(0 ) Cx1 0,y'x(0 ) Cx2 1 yx(t) e tsint2.求yf(t)

t 0

0

0

首先确定y'f(0 )与yf(0 )

0 0

0

0

0

0

yf''dt 2 yf'dt+2 yfdt= (t)dt

可得y'f(0 ) y'f(0 ) 1,yf(0 ) yf(0 ) 0; y'f(0 ) 1则 yf'' 2yf' 2yf (t)

y(0) 0 f

当t 1时，yf'' 2yf' 2yf 0 yf(t) e t(Acost sint)

代入初始条件：y'f(0 ) B 1,yf(0 ) A 0 yf(t) e tsint (t)3.求全响应y(t)y(t) yx yf 2e tsint

t 0

2.4 （1）y(k+2)+3y(k+1)+2y(k)=0,yx(0) 2,yx(1) 1 解：特征方程r 3r 2 0 (r+1)(r+2)=0 特征根： r1 1,r2 2

y(k)=Cx1r1 Cx2r2 Cx1( 1) Cx2( 2) 代入初始条件

k

k

k

k

2

Cx1 Cx2 2

解得Cx1 5,Cx2 3

C 2C 1x2 x1

yx(k) 5( 1)k 3( 2)k k 0

(2)y(k+2)+2y(k+1)+2y(k)=0. 解：

yx(0) 0,yx(1) 1.