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）

d [(a+bx)c ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
7 ^( ^5 @7 U& }% I# X1 q* O
多谢了！
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% j1 f. L& o! g% f[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s

3 [) m5 p3 J' f! j(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
; J' D& J" [3 {$ T(a+bx+kx)c=sx
7 J) K% l1 ]& Y& K$ ]9 R  Pc=sx/(a+bx+kx)

十年移民路_

Solution:
  `0 v) h- @. @8 T7 x. B
, P/ h$ a! U/ X! y+ \0 b9 ~' E  OFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
- n( Z' k. A8 n' [( M1 h% \
(a+bx) dC(x)/dx  = -(k+b)C(x) +s


introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
% i0 D# Z. V4 }# Gwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
- T+ p& f2 G# P; z# btherefore:
% G6 ]# E% m/ Y& n# t! f4 h) c9 f0 f
  s* K( R; L4 I. w{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:
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( ]& Q- g3 D: MdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

so that:   ln Y(x) =( K/b) ln(a+bx)
' ?9 f$ S' W9 X1 X3 a. k
this means:  Y(x) = (a+bx)^(K/b)
4 n' a1 }4 V0 U0 ?+ i! V9 gby using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)

) \1 i; w+ d( Sfinally:
' i( y' j$ Y( e1 ]1 Q
% z+ X" G4 N% S7 t& e1 f& qC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)