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）

* V' H& h* Z; l7 s! ^: y, Vd [(a+bx)c ] /dx = -kc +s
, \4 \3 K: {9 [7 M. b* Twhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

* l7 o& b$ w, w' U% z# M: |多谢了！
) s0 O. _5 {, n
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s

$ i/ C: [5 `6 G(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
3 y8 L6 R5 E8 W# V! g5 p8 N  g(a+bx+kx)c=sx
3 F- V, S3 D7 Y! G( [; i9 fc=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

# X$ f' s& y: n8 t(a+bx) dC(x)/dx  = -(k+b)C(x) +s
3 e# k3 Y' t( y- R3 n  O; o
; e2 R+ Z" G8 l, y! @
! D* \7 P" K" t7 P$ X1 Nintroduce 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)

from here, we can get:
! W: v# K7 j/ `5 q; @
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

, o+ q+ `+ x9 m+ }  }so that:   ln Y(x) =( K/b) ln(a+bx)
: h4 _" s* @4 b; r- k" i! W
8 v0 R  l& Z4 athis means:  Y(x) = (a+bx)^(K/b)
, Z7 {7 V9 u  j) Wby using early transform, we can have:
, z3 u, ~& t7 u6 q( U5 Z1 H: d
-(k+b)C(x)+s = (a+bx)^(k/b+1)
1 o4 r1 w# |* z5 j
- G: |1 ~" b' e& sfinally:

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