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[ÇóÖú] MATLA BÇó½â΢·Ö·½³Ì ÒÑÓÐ1È˲ÎÓë

U=[1 2 3;4 5 6;7 8 9];
[a,b,c] = dsolve('Da = -(a*U(1,1)+b*U(2,1)+c*U(3,1),Db = -(a*U(1,2)+b*U(2,2)+c*U(3,2),Dc = -(a*U(1,3)+b*U(2,3)+c*U(3,3)','a(0)=1,b(0)=0,c(0)=0','t')

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wurongjun

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ÓÐÀÛÈË1: ½ð±Ò+3 2015-06-18 14:55:43
¿ÉÒÔ½â,²»¹ýÒª°ÑUÀïÃæµÄ¾ßÌåÊýÖµ´øµ½ÀïÃæ!
ÒòΪÕâÊǾ«È·(·ûºÅ)½â,µ¥ÒýºÅÀïÃæµÄ°´±í´ïʽ¿¼ÂÇ!
>> [a b c]=dsolve('Da = -(a*1+b*4+c*7),Db = -(a*2+b*5+c*8),Dc = -(a*3+b*6+c*9)','a(0)=1,b(0)=0,c(0)=0','t')
a =
2/33*(22+55*exp(3/2*(-5+33^(1/2))*t)+7*33^(1/2)*exp(3/2*(-5+33^(1/2))*t)-7*33^(1/2)*exp(-3/2*(5+33^(1/2))*t)+55*exp(-3/2*(5+33^(1/2))*t))/(5+33^(1/2))/(-5+33^(1/2))
b =
4/99*(33*exp(3/2*(-5+33^(1/2))*t)+33^(1/2)*exp(3/2*(-5+33^(1/2))*t)-66-33^(1/2)*exp(-3/2*(5+33^(1/2))*t)+33*exp(-3/2*(5+33^(1/2))*t))/(5+33^(1/2))/(-5+33^(1/2))
c =
2/99*(-33*exp(3/2*(-5+33^(1/2))*t)-17*33^(1/2)*exp(3/2*(-5+33^(1/2))*t)+66+17*33^(1/2)*exp(-3/2*(5+33^(1/2))*t)-33*exp(-3/2*(5+33^(1/2))*t))/(5+33^(1/2))/(-5+33^(1/2))
>>
ÑéÖ¤Ò»ÏÂ!

>> subs([a b c],t,0)
ans =
    1.0000         0   -0.0000
>>
ÉÆ¶ñµ½Í·ÖÕÓб¨,È˼äÕýµÀÊDz×É£.
2Â¥2015-06-16 21:51:51
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

ÓÐÀÛÈË1

½ð³æ (³õÈëÎÄ̳)

ÒýÓûØÌû:
2Â¥: Originally posted by wurongjun at 2015-06-16 21:51:51
¿ÉÒÔ½â,²»¹ýÒª°ÑUÀïÃæµÄ¾ßÌåÊýÖµ´øµ½ÀïÃæ!
ÒòΪÕâÊǾ«È·(·ûºÅ)½â,µ¥ÒýºÅÀïÃæµÄ°´±í´ïʽ¿¼ÂÇ!
>> =dsolve('Da = -(a*1+b*4+c*7),Db = -(a*2+b*5+c*8),Dc = -(a*3+b*6+c*9)','a(0)=1,b(0)=0,c(0)=0','t')
a  ...

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