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?
  @* G+ w2 @" Z" a
" s3 z3 |6 Q4 `: {多谢了！

1 r8 ?: c8 f9 u9 k. T8 _[ 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
, u# w) Q+ |" A$ i% Y; K* c  w# ?ac+bxc=-kxc+sx
, ^& a+ o4 O0 j2 E/ U! `(a+bx+kx)c=sx
4 R& x( E, K2 i0 {/ r% M; Dc=sx/(a+bx+kx)

十年移民路_

Solution:
) x! `# S* a# x9 t/ I0 x( u
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
- c0 D8 j+ d. z- s3 \& bso:
% {/ q) f8 ~( L: \6 \5 \) I
" e- S) Z' o- NbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
3 v& l, G- h& M% j9 Y6 bi.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s


% D6 w5 u% `# hintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
6 B4 m* q% ?; `& o( I6 ~- `. u. K& swhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

{(a+bx)/K} dY(x)/dx=Y(x)
4 e: }$ ]- k$ n9 H
; z, @2 v5 ?1 K) T' C# \0 O$ ]from here, we can get:

4 J$ o4 W9 x) ydY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
* g. c6 n  N& P( o$ K) a# ?
" X  }* G4 x5 ], e- {( jso that:   ln Y(x) =( K/b) ln(a+bx)
% F' s: j/ h5 k0 S# L0 k
& t, x) B7 ?' v; J2 q9 k$ gthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
4 f/ J4 X! K! k9 d2 W
-(k+b)C(x)+s = (a+bx)^(k/b+1)
7 H6 G/ \9 F9 Z. X0 l
$ k) ^0 q2 H, e, w# {, R% S  xfinally:

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