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
where: 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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1 T+ J1 k8 E! l# e# `/ b/ y[ 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 ?$ o- R0 y1 T% S8 Q(a+bx)c  = (-kc +s)x
5 s3 a- N" @! D3 Kac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
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- X9 J: Z. b# yFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
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4 z5 T& |# l7 b& C3 M2 B, v! t. IbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

: j0 _7 y- b0 Y5 `2 g" |2 K3 U* ?(a+bx) dC(x)/dx  = -(k+b)C(x) +s
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* E5 U) B# Z, V9 ~/ |; S$ f& Ointroduce 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
: [9 O$ h0 F& V5 Y9 Y2 s2 ctherefore:
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: k; B( A% }4 v- z  h  O! u{(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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' C0 _6 r2 a5 d% nso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
6 E- c1 h% x% L9 Y9 n; W- yby using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)

4 ~6 G- ]+ o* @- P) B: Lfinally:
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' p7 d. _3 H2 k: dC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)