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）
% u' @) u* y7 W1 \; o
d [(a+bx)c ] /dx = -kc +s
. c6 S+ ^3 R3 D* h# {! Vwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
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! X2 j) z# e8 {4 n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
3 o7 |6 H! n6 Y
* u( y1 s! H4 C- H(a+bx)c  = (-kc +s)x
$ S7 f# c: b6 n( Y+ U9 uac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
4 ]/ _+ J. |) z! r9 ~" K
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
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6 |  o8 T8 G# x0 N4 z; KbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
- [0 |9 m$ j5 L5 L+ {
0 A; q) n; }8 N7 z- k3 b(a+bx) dC(x)/dx  = -(k+b)C(x) +s

( g9 g1 i  k* V$ z0 m4 j: r# T4 B& X
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
. Y' n) n# p4 a" Cwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

; M& ^/ c0 G* v' R! S# K{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:

3 w5 H" `, N+ R9 q7 W# d- E% JdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

/ c# P$ D. q0 w8 Yso that:   ln Y(x) =( K/b) ln(a+bx)
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' c2 M; K/ c2 i1 B, e: Nthis means:  Y(x) = (a+bx)^(K/b)
/ {" j3 M! [. i5 L# E4 yby using early transform, we can have:
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' Q' }8 Q. o. [9 L% Y-(k+b)C(x)+s = (a+bx)^(k/b+1)
3 u* V1 |9 s/ z1 g
finally:
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8 K9 A/ B& {& U8 ?: J& v) gC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)