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）
9 s7 o  o! x" b7 E' d6 l: z; V6 P
2 S8 c" @7 A6 Y+ M7 A7 ud [(a+bx)c ] /dx = -kc +s
+ u: d9 j; m7 B' K( ^where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
/ {6 v" J) J6 Y! [$ G' D
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

9 h) S) k# R0 `9 @(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
" i# W( i, j( H9 y0 z& m8 w(a+bx+kx)c=sx
# [8 y' c/ m+ W* Uc=sx/(a+bx+kx)

十年移民路_

Solution:

" m, B: {3 w% \: v% D1 MFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
! p$ \8 Z6 V5 S! D9 Mso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
5 N  P/ ?- t9 M8 ~3 A; Ai.e.
- T9 m4 y4 L8 T0 R( N# Q5 j7 D; q  X
- x8 F8 W( e( F6 [(a+bx) dC(x)/dx  = -(k+b)C(x) +s
9 M* @0 r, \9 Z; r& k2 }9 s6 w
. \1 P+ {3 y' U3 @- ^( R# E/ R
; o  B. a9 Y! p# w9 N% m, Jintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
3 a9 U/ g0 r6 c" vwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
$ ?. T9 ^5 A0 u2 W" ftherefore:
% F) y3 \3 e7 E7 T' d6 P
{(a+bx)/K} dY(x)/dx=Y(x)
  y" j9 E$ |& J
+ R. X5 p& o" Zfrom here, we can get:
9 b. }' j- H; a! d5 T3 Z7 p
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
9 _0 T" r" H. J7 o" Z( d" D
" G% c/ b. l1 c- Nso that:   ln Y(x) =( K/b) ln(a+bx)
" \: u) [. B6 y, Y8 a5 {( j( m2 l
this means:  Y(x) = (a+bx)^(K/b)
) }4 M. V0 @" i2 X) kby using early transform, we can have:
1 ], U( e0 l& B3 V
-(k+b)C(x)+s = (a+bx)^(k/b+1)
6 b8 I' b- R2 _% D
# Y5 w. k  w7 j# E" Jfinally:
) n5 F, c, N0 i: Y
6 B/ I& V% E; Y( x/ a2 GC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)