[y1'] [ 1  1 1 ][y1]
[y2']=[ 1  0 1 ][y2]  , y1(0)=1, y2(0)=1, y3(0)=-1
[y3'] [-1 -1 -1][y3]

y1'=y1+y2+y3

y2'=y1+y3

y3'=-(y1+y2+y3)

=>

 (y1+y3)'=0

又

 y2'=y1+y3

所以

y2''=0 => y2=c1x+c2 ...(1)

 y1'=y1+y2+y3

    =y2'+y2

    =c1x+c1+c2
=>



 y1=0.5*c1x^2+(c1+c2)x+c3 ...(2)

因為 y2'=y1+y3

所以 y3=y2'-y1

       =-0.5*c1x^2-(c1+c2)x+c1-c3...(3)

得

 y1=0.5*c1x^2+(c1+c2)x+c3

 y2=c1x+c2

 y3=-0.5*c1x^2-(c1+c2)x+c1-c3

且

 y1(0)=1, y2(0)=1, y3(0)=-1

故

 c3=1 ,c2=1, c1-c3=-1

=>

 c1=0 ,c2=1 c3=1

=>

 y1=x+1

 y2=1

 y3=-x-1


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※ 編輯: jack0711        來自: 140.113.123.237      (02/24 02:14)
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