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）
. c- q/ h- N) X+ @. n6 w
d [(a+bx)c ] /dx = -kc +s
  }3 I4 t: P: H3 R* n8 N" Q% nwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！

" `6 ~9 R# H1 O" d% x2 K+ e# D[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
3 `: v4 J0 W6 B, d4 S  U
(a+bx)c  = (-kc +s)x
! M8 S: W! e: p2 Z# w9 Gac+bxc=-kxc+sx
4 T  g! k5 B! g9 p" ^(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
4 [2 k, ?1 H6 q8 b4 @: C" K
% ^1 e6 d7 K$ ~. M& m" P( ^From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

6 x1 S1 I( r; F: I(a+bx) dC(x)/dx  = -(k+b)C(x) +s
, f. r9 ]- Q7 a% i5 g, `. `8 W! A  \

( R3 X) X# V' \4 @introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

( i) Q" i3 ^) \) N9 t2 z* }8 G{(a+bx)/K} dY(x)/dx=Y(x)
) M, v0 a& q6 o
( g4 E2 m  _1 h# T4 y) V$ Xfrom here, we can get:

+ j  Q8 q" V; r5 t+ ~dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
7 s% q8 O  b" l% C1 E
  ~, b0 o2 }3 c( a( o- ?% D( ~so that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
* Z  c5 J4 n/ u' T8 V3 |  ^4 x) G
7 Z1 ]$ l1 g% v-(k+b)C(x)+s = (a+bx)^(k/b+1)

$ I* {* f2 Y6 b+ g: Mfinally:
/ ?8 G; a0 ~) Q% k( k5 q
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)