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）
2 ]3 y1 `% E0 a
8 q  C) S* U  p! F7 W, Td [(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?

多谢了！
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- m! `: ^5 v" H; L9 W$ \[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
- b$ w6 R, ]# o8 S5 D+ }
2 \1 Z8 a6 ~, Z- N) U8 ~- Z) X(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
3 I2 D7 j+ q" ?$ C# w5 D$ o(a+bx+kx)c=sx
4 b3 W2 ?/ E+ |" u3 H( q# pc=sx/(a+bx+kx)

十年移民路_

Solution:
  Z; {$ \2 A1 R# K
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
5 v/ n# |: \$ f9 k
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

- f8 \5 ~; I/ X: B: `. O(a+bx) dC(x)/dx  = -(k+b)C(x) +s


introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
6 G5 N  y! M$ `8 }which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
" J* U6 ]/ x- X. ?  P( y4 R- [
  k1 h2 r" O' B4 n6 }% d, Q, C{(a+bx)/K} dY(x)/dx=Y(x)
1 G0 D, F( m  e+ ^- U
from here, we can get:

dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

1 G8 J" s6 Y! q9 c1 |9 w& t/ G* S* Wso that:   ln Y(x) =( K/b) ln(a+bx)

2 \0 S9 X7 M5 i* Bthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

" A4 S! s: Q1 K$ U5 A6 t-(k+b)C(x)+s = (a+bx)^(k/b+1)

* \4 ]) E& C% [, a( h  f& A5 nfinally:
; g( j* H  ?8 N& L
4 M0 y5 ]- Z7 S' d4 nC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)