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）
  j3 B7 o) {6 Y0 }! }
d [(a+bx)c ] /dx = -kc +s
/ Q" e9 L. c: O1 T% xwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！
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6 n" V' {" _# K# C5 O+ k[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
- F+ I/ |2 i* y0 \. e- ]6 u; h% n
. E" R0 K6 M' \9 O5 U6 f" r(a+bx)c  = (-kc +s)x
, s( O3 S- ^3 J3 ?ac+bxc=-kxc+sx
( S5 U: W) I* G3 Z( t(a+bx+kx)c=sx
5 M, j- L" d- R' P6 jc=sx/(a+bx+kx)

十年移民路_

Solution:
7 s6 K! o( E8 O! ~
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

- j# H: N" J5 l5 @# t9 {, {bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
/ t8 d5 {: v$ X2 u# \% `i.e.
# T5 x6 E0 f3 [+ ~* U# g) T
(a+bx) dC(x)/dx  = -(k+b)C(x) +s

  g$ [* r. ?: |5 r- Z- V  t
6 {5 x7 s* i. Q& V6 Aintroduce 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
; v. q; K0 r: \# P! h: F/ ]0 t* X7 Vtherefore:

{(a+bx)/K} dY(x)/dx=Y(x)
6 z, ?7 T* ]2 k9 R6 r
from here, we can get:

: v$ F# g) w- z/ E7 w. ^. hdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

, U9 S  D& x9 F( Dso that:   ln Y(x) =( K/b) ln(a+bx)
3 N8 l+ L: |0 x3 h' Q; |. g0 C. v! U
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
6 |+ T/ i9 Z( i/ O6 k7 N! k
; v' C) y& L7 O( v( v-(k+b)C(x)+s = (a+bx)^(k/b+1)
& Y1 b$ M% ]+ S% d/ {
finally:

C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)