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）
" m# l6 E+ h! G9 {! t+ l) T9 \7 Z
, }0 i  I: A3 E  R' W% ?' n9 ~d [(a+bx)c ] /dx = -kc +s
% D1 O+ u; U) |; Ewhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

$ |( e4 Y  ]' {6 g多谢了！

3 b/ P/ U) b9 G8 m7 g$ m[ 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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(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
* n2 |1 J4 s+ X) _c=sx/(a+bx+kx)

十年移民路_

Solution:
) g$ z  x) b4 X0 }  X
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
5 S- H! r* u# G' t% uso:

7 i, T' m5 ?( o9 A7 b9 gbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
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+ t$ ~5 J3 m2 j' L(a+bx) dC(x)/dx  = -(k+b)C(x) +s
. ?: n6 b% k) f+ V, ^8 l+ _0 y
" r5 Q1 R5 T0 S* B8 [3 j/ s8 q# l
introduce 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
therefore:
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{(a+bx)/K} dY(x)/dx=Y(x)

, X( o+ [1 t: j* f5 C6 qfrom here, we can get:

% l& x5 K9 |' }1 q0 C1 z0 `dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
8 R! c% y: x/ w0 h$ \
so that:   ln Y(x) =( K/b) ln(a+bx)

( k6 O! [6 H' C0 Ythis 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)
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  F* c6 q. e9 U! |3 Q$ j, C2 Rfinally:

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