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）
% `/ ?' c5 B- r2 s8 o( b
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?
+ C8 @' H/ E- V- r; A1 r; O/ {6 ]- g
多谢了！

[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
5 d8 I6 e) R+ E2 r$ W. x( _/ C
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
# \4 h! h% q% `, D9 q# C(a+bx+kx)c=sx
/ j4 S! b! A$ k" Pc=sx/(a+bx+kx)

十年移民路_

Solution:

% h7 d2 l& o4 Y$ zFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

3 _) o0 N  h- J) }* zbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
* x  B* S! D% o1 Q+ ^
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
# V1 P; V9 d0 \* s

introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
; E- |' W6 ^: w. Y) S/ y' ~/ ywhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
, a3 D+ @/ f2 g/ P4 Z$ mtherefore:
9 M: Q% p* w8 n& p; b- k5 I
# K# S0 Q3 _6 y8 _2 d0 r{(a+bx)/K} dY(x)/dx=Y(x)
2 K) D9 [% v4 {' f) u5 F0 i
9 Q3 N: c8 @! m; l5 Yfrom here, we can get:
% V# J1 |6 K$ {: w( [; d4 S# [
, v9 r6 G  C" Q# U4 y2 V. cdY(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)
; E$ r9 A, M3 B9 c! B, c
0 Q& p/ n1 a/ `7 Ithis means:  Y(x) = (a+bx)^(K/b)
" Z$ z. \7 U: o: R) v. O* ~by using early transform, we can have:
1 E! H( }& P, _6 w' J
/ [0 @5 r2 @2 n4 q7 P% Z. @-(k+b)C(x)+s = (a+bx)^(k/b+1)
9 p) @) h( n- k$ D
finally:

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