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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0 |" `6 ~; L- Q: C* o$ c; R8 \* ?  od [(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?
# W+ D' N, v9 R  u1 H9 y8 Y
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
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2 i- U) z. a2 u( O7 z9 D(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
& g. x" u( d- ]0 @(a+bx+kx)c=sx
c=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
so:
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bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s

0 @1 ~, Q6 f; K. g7 c4 r- ^" L  j
* n5 L* {) {; e' H9 U2 Lintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
& Y& c7 h! }% J# y# L/ l. l( cwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
- Y, h/ ]. E  ]: z% N. ytherefore:

/ D, |: A! k8 ^# _9 x{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:

dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
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) ~3 G" j# l2 Z5 u8 V! H' ?$ G2 [' Xso that:   ln Y(x) =( K/b) ln(a+bx)
5 H7 _6 R! l2 ^" n# W' A6 X
this means:  Y(x) = (a+bx)^(K/b)
6 p( ?7 _3 E5 @5 c! |( C: p( e: ~by using early transform, we can have:
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-(k+b)C(x)+s = (a+bx)^(k/b+1)
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4 T1 ^& C' E6 m' }3 E+ d6 Yfinally:

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