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dy(1)/dt=(a+b+c)*y(1)+g*y(2)+b*y(3)
dy(2)/dt=-(a+b)*y(3)
dy(3)/dt=(a-b-c)*y(3)-g*y(1)+2*g*y(2)*y(1)
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jjdg(½ð±Ò+1): ¸Ðл²ÎÓë 2011-11-18 01:05:11
ÒýÓûØÌû:
3Â¥: Originally posted by koorb at 2011-11-17 09:46:25:
ÏÂÃæÊdzÌÐò´úÂ룬Óï·¨¿Ï¶¨Ã»´í£¬Ö»ÒªÄãÔÙ¸ø¶¨³õʼÌõ¼þ£¬¾Í¿ÉÒԵõ½½á¹û¡£·ñÔò½â³öÀ´µÄÊÇͨ½â¡£
eq1 = y1'[t] == (a + b + c)*y1[t] + g*y2[t] + b*y3[t]
eq2 = y2'[t] == -(a + b)*y3[t]
eq3 = y3'[t] == (a - ...

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5Â¥2011-11-17 16:25:28
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jjdg(½ð±Ò+1): ¸Ðл²ÎÓë 2011-11-17 02:41:04
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2Â¥2011-11-17 00:30:03
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koorb

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ÓàÔó³É(½ð±Ò+2): лл²ÎÓëÓ¦Öú£¡ 2011-11-19 14:28:38
liuhaizhen(½ð±Ò+10): ¸ÐлÄãµÄÓ¦Öú 2011-11-28 10:00:33
liuhaizhen(½ð±Ò+5): ÓÅÐã 2011-12-07 10:16:41
ÏÂÃæÊdzÌÐò´úÂ룬Óï·¨¿Ï¶¨Ã»´í£¬Ö»ÒªÄãÔÙ¸ø¶¨³õʼÌõ¼þ£¬¾Í¿ÉÒԵõ½½á¹û¡£·ñÔò½â³öÀ´µÄÊÇͨ½â¡£
eq1 = y1'[t] == (a + b + c)*y1[t] + g*y2[t] + b*y3[t]
eq2 = y2'[t] == -(a + b)*y3[t]
eq3 = y3'[t] == (a - b - c)*y3[t] - g*y1[t] + 2*g*y2[t]*y1[t]
DSolve[{eq1, eq2, eq3}, {y1[t], y2[t], y3[t]}, t]
ÌìµÀ
3Â¥2011-11-17 09:46:25
ÒÑÔÄ   »Ø¸´´ËÂ¥   ¹Ø×¢TA ¸øTA·¢ÏûÏ¢ ËÍTAºì»¨ TAµÄ»ØÌû

koorb

Ìú³æ (³õÈëÎÄ̳)

¡¾´ð°¸¡¿Ó¦Öú»ØÌû

ÒýÓûØÌû:
3Â¥: Originally posted by koorb at 2011-11-17 09:46:25:
ÏÂÃæÊdzÌÐò´úÂ룬Óï·¨¿Ï¶¨Ã»´í£¬Ö»ÒªÄãÔÙ¸ø¶¨³õʼÌõ¼þ£¬¾Í¿ÉÒԵõ½½á¹û¡£·ñÔò½â³öÀ´µÄÊÇͨ½â¡£
eq1 = y1'[t] == (a + b + c)*y1[t] + g*y2[t] + b*y3[t]
eq2 = y2'[t] == -(a + b)*y3[t]
eq3 = y3'[t] == (a - ...

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