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8Â¥: Originally posted by peterflyer at 2013-11-07 01:11:03
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11Â¥2013-11-07 13:59:58
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9Â¥: Originally posted by zywang1999 at 2013-11-07 10:01:07
n=0ʱֱ½ÓʹÓÃÒ»½×ÏßÐÔ΢·Ö·½³ÌÇó½â¹«Ê½µÃ: c = k * exp(at)  - b/a; ÆäÖÐkΪÈÎÒâ³£Êý,ÏÂͬ
n=1ʱc'=(a+b)*c, c'/c = a+b, c=k*exp.
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12Â¥2013-11-07 16:31:26
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10Â¥: Originally posted by peterflyer at 2013-11-07 13:59:22
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dC/dt = a*C+b*C^n£¬a£¬b£¬nΪ³£Êý£¬Õâ²»¾ÍÊDz®Å¬Àû·½³ÌÂð¡£Â¥Ö÷¿´×ÅÎÒÔõô½âѽ£º
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  C^(-n)*dC/dt-a*C^(1-n)=b,  ¼´£º*dC^(1-n)/dt-a*C^(1-n)=b
   Áîu=C^£¨1-n£©,£º*du/dt- ...

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13Â¥2013-11-07 16:33:07
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13Â¥: Originally posted by fengweiche at 2013-11-07 16:33:07
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14Â¥2013-11-07 17:32:31
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13Â¥: Originally posted by fengweiche at 2013-11-07 16:33:07
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fengweiche: ½ð±Ò+7, ¡ï¡ï¡ï¡ï¡ï×î¼Ñ´ð°¸, ·Ç³£¸Ðл1 2013-11-14 20:09:48
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dC/dt=a*C+b*C^n=b*C*[C^(n-1)+a/b]
dC/{C*[C^(n-1)+a/b]}=b*dt
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1/{C*[C^(n-1)+a/b]}=a1/C+{a2*C^(n-2)+...+an/[C*[C^(n-1)+a/b]
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     a1=b/a,  a2=-b/a,   a3=a4=...=a(n-1)=an=0
   b/a*dC/C--b/a*C^(n-2)*dC/[C^(n-1)+a/b]=b*dt
   b/a*dC/C-b/[a*(n-1)]*d[C^(n-1)+b/a]/[C^(n-1)+a/b]=b*dt
   b/a*lnC-b/[a*(n-1)]*ln[C^(n-1)+b/a]=b*t+A     ,AΪ»ý·Ö³£Êý¡£
  Õâ¾ÍÊÇCÓëtµÄº¯Êý¹ØÏµ¡£¿ÉϧµÄÊÇÖ»ÄÜд³ÉÒþº¯ÊýµÄÐÎʽÁË¡£

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