请教大家一道微分题，多谢！(原题贴错了，现纠正）

醉酒当歌

请教大家一道微分题，多谢！(原题贴错了，现纠正）
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d [(a+bx)c ] /dx = -kc +s
0 s9 l+ L* L/ @% O1 v4 m9 hwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
7 ~+ I$ Z4 w- f% s& Y5 K" c
多谢了！
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3 W; k  X$ y" Z[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
  ], J/ c" k0 o# w6 r: J. F% R
5 h9 a( s& R7 v  X$ E; D(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
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From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
3 h. P, \5 C# U6 v' S3 n' \4 oso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
+ G( ]& ]) }  o
(a+bx) dC(x)/dx  = -(k+b)C(x) +s


1 H9 G6 `6 `! ?) V8 @, Y/ Tintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
+ _' _3 v% `) k  f( owhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
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! \8 _* S$ T1 K{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:
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$ c# w4 Q# k! q$ [. hdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
$ n4 y0 f: G; u0 ]
so that:   ln Y(x) =( K/b) ln(a+bx)
  B7 b2 }) f+ Z  F% Q
1 L4 u3 `$ x6 qthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
2 U' b' S/ ^* A0 g2 x( ~
-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
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- D" z& Y, q4 O3 g$ p& a! O8 bC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)