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）
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d [(a+bx)c ] /dx = -kc +s
- D1 p0 l8 K) e% x8 wwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
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多谢了！
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[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

(a+bx)c  = (-kc +s)x
( m, G8 z# {) xac+bxc=-kxc+sx
(a+bx+kx)c=sx
" d6 _6 d) s/ A2 L( S" Vc=sx/(a+bx+kx)

十年移民路_

Solution:
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From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
  R1 w. [( u& q# @. X, t4 pso:
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7 U# ~2 \; b7 L% L+ `bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
# O# @! S" m5 ?$ V4 W4 Si.e.
% e/ u+ ?: r* J+ G! K2 x
! k* o. [& }# n3 ?' T& g! J(a+bx) dC(x)/dx  = -(k+b)C(x) +s
$ f0 q# @3 W9 O& p) j8 x
1 |7 ?% t8 R1 C$ `; a  |
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
( t$ Q3 u" l% L5 x( g  ]8 Vtherefore:
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1 H( w4 E, S" j9 J% ~{(a+bx)/K} dY(x)/dx=Y(x)
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( @+ {! E3 h; W& cfrom here, we can get:
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+ k; _& w0 q6 @0 O" f7 P; MdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

1 ~  o! {& h  Cso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
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% V) G( R4 [$ z, O-(k+b)C(x)+s = (a+bx)^(k/b+1)

: u2 A" E+ P/ h& \  Ufinally:

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