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）
. A7 k7 s' n( [. W% C* O6 {
d [(a+bx)c ] /dx = -kc +s
6 Z+ {2 j/ W& e* e" d5 Ywhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
9 \0 b2 n$ z) y% q
" Q0 o0 i  I( M多谢了！

. g7 i6 q& V5 }$ W" ^) {1 `[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

(a+bx)c  = (-kc +s)x
. [- `! o& V' g9 H$ z3 @5 U8 q0 yac+bxc=-kxc+sx
& {0 b/ P. Q5 ~1 J8 ^) F(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
  b1 m' p  c( c- b9 P+ w
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

; p, E1 k  s) r& \# o8 O! abC(x) + (a+bx) dC(x)/dx = -kC(x) +s
) W. v" t! o4 E0 D- s3 \! }, O/ m# Yi.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s

6 U. Y5 |. R$ O/ d8 T- q
, U0 T7 _8 M9 a% \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:

* Y. \7 X* i# B: O{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:
" x* q: O4 R6 `
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
) R7 }6 ?4 R, l: }$ d; N, U
so that:   ln Y(x) =( K/b) ln(a+bx)

0 s% `8 s6 V) n% `! [' bthis means:  Y(x) = (a+bx)^(K/b)
4 M* n! r5 g6 J. h: B* Yby using early transform, we can have:
3 `2 P' u% U: q; g- \' @
+ w9 \* m% w4 f% P0 S% [9 b-(k+b)C(x)+s = (a+bx)^(k/b+1)
+ f; r/ s0 I5 S) o
- v/ p. D/ k7 sfinally:
4 s4 F+ m; d3 N0 e# e" J2 T
8 y. _: T# I. wC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)