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）

d [(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?
+ {& N% ]! ^9 A, Y/ I
多谢了！
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3 X7 @! g0 e/ B. J' r3 [8 A[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
5 s4 I2 |1 M! K- L0 I! R9 Q
(a+bx)c  = (-kc +s)x
! V; f/ }$ C1 `ac+bxc=-kxc+sx
, M: ?& x  g1 D) Y/ C! w(a+bx+kx)c=sx
1 e& M* m2 g( r' L# q8 O' }c=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
* [" `, ]( t" n9 D  h
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
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(a+bx) dC(x)/dx  = -(k+b)C(x) +s


- L9 S! ?" S! N6 E9 \' @# Lintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
. A& _, W! i: u0 |/ R! Zwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
- p. r- ^0 Y  Z1 J% Etherefore:

: t' H3 ~+ _* ~2 N{(a+bx)/K} dY(x)/dx=Y(x)
! n/ a1 p8 ]/ b. D# y$ }/ X
$ s4 z$ Y& `$ q- n- @from here, we can get:
: _4 a' @# {2 o
# U5 I' D" {& T6 J0 y; O7 OdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
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so 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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+ y$ X$ k& D  N( g-(k+b)C(x)+s = (a+bx)^(k/b+1)

  v/ ]6 u. G1 \4 S* ~( E: u8 D! efinally:

/ ^) y! p1 c# V3 v. sC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)