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）
: W7 d! c  l% s) m& S. ~' u
d [(a+bx)c ] /dx = -kc +s
* f$ S8 T0 C4 ^  n5 L9 N, }* {where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
. K7 `% R4 \* a! e) N& e
+ U- I9 R4 N7 \6 T1 x/ ?多谢了！
; j, u. \, \% B1 s  z! ~
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
! Q1 b" u5 n$ R
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
- S- N" S) E. E' m# Wc=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
$ d  X- ]: A: C( [( ^so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
: I0 n" A- U; m  U
(a+bx) dC(x)/dx  = -(k+b)C(x) +s

  W0 ^4 ]" N5 x: p2 J  M8 Q
; e+ F6 s+ t# Z/ D$ n% r* pintroduce 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:

{(a+bx)/K} dY(x)/dx=Y(x)
2 x+ I# {! U7 Z$ ^
from here, we can get:
0 w% E1 N5 d! y  F4 f
) N0 O5 b# g/ `+ ^5 W( YdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
- P  n* Z; s( c- [+ R' ~* P& D$ Q
9 r5 B  g8 n) k' Sso that:   ln Y(x) =( K/b) ln(a+bx)

9 k5 t6 V* c0 h5 I9 ], X- _this means:  Y(x) = (a+bx)^(K/b)
  e; l$ p9 q1 E. I5 C! rby using early transform, we can have:

, o, R/ A. [, X+ z4 C: R-(k+b)C(x)+s = (a+bx)^(k/b+1)
/ G( J) `1 q4 s5 H' D: ?4 H9 V
! z) e! J1 @* Y+ n+ F1 n! |; Gfinally:

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