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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' ^& K" j' H' ~3 f, g: `9 x7 P5 dd [(a+bx)c ] /dx = -kc +s
) a) |& S( _$ v, f3 Z9 Fwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！
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5 T1 }0 d1 T9 v7 o9 s3 M( G0 L[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
6 `8 l- C/ c9 f3 g: Z2 @% I: y( K
. t) V6 [* H0 i(a+bx)c  = (-kc +s)x
2 l) u7 u- }; n* T4 N. k; Oac+bxc=-kxc+sx
(a+bx+kx)c=sx
4 |9 \+ F& T( S( {& Pc=sx/(a+bx+kx)

十年移民路_

Solution:
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! ]* D: o! ^) {/ `From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
, R: e; Y( ?, w8 T) ~- dso:
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0 `- q2 i" t4 b3 d# K8 |bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

$ w' @2 e2 E. Q* H. U(a+bx) dC(x)/dx  = -(k+b)C(x) +s


' }& L/ m- O, {8 cintroduce 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
% a% z2 I) u0 `  R2 L3 rtherefore:
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/ s+ |; N& @5 a{(a+bx)/K} dY(x)/dx=Y(x)

& k' X" m/ T8 A  g3 N# dfrom 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)
" ~4 |$ k  o; D1 \& ~# d
so that:   ln Y(x) =( K/b) ln(a+bx)
5 z; ^( }/ Y; Y
; c1 d$ O- |6 G6 }7 ithis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
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-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

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