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）

& t( F( w- p' V3 h) ]% Pd [(a+bx)c ] /dx = -kc +s
7 P& p' p& m3 o* q. P  T' B! _where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
/ j% ^3 v3 V& h2 T# M8 S6 q
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
$ p5 d% ~$ e4 \* x' g! _  f
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
+ D  h" l9 {( [' w" f(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
3 h- M, F6 B8 I$ f  z: C2 Cso:
: ~: t8 g3 h6 x' q
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
: ?- v3 V! f8 ?* bi.e.

3 k: v' g: m7 k  y( Z  e(a+bx) dC(x)/dx  = -(k+b)C(x) +s
& x, P8 {2 N$ b6 t- Q
; b4 }+ m9 p% N5 |
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
, H! M- C& B* P* htherefore:
( q# `0 Z. @* N9 Z" E# `
- C9 [- Y/ }; e7 G$ n{(a+bx)/K} dY(x)/dx=Y(x)
4 V6 x; f2 ?) r$ j7 ?: x7 M
from here, we can get:
- T) A. d% O% s; Y- E1 Y( E
. c8 Z, }' m+ g0 d3 \+ A+ EdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
0 e, n' ]; F( o) |" b
# y4 y7 y' t- m# W" Q% R6 \" }so that:   ln Y(x) =( K/b) ln(a+bx)
, K) ?/ _, k* ^: B7 c6 m8 h
# I, r. z$ w7 w9 Othis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

9 z; ?2 g% l% O" B$ I& R9 g2 D/ |-(k+b)C(x)+s = (a+bx)^(k/b+1)
; b" `+ g& r+ L! N6 x, B
finally:

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