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！

- v! S" S4 d& [6 Y, [" n( J[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
1 R( g+ h; \: o- P0 ?
0 A' U& U: ~" b(a+bx)c  = (-kc +s)x
3 e# r+ g' L- F8 uac+bxc=-kxc+sx
0 ]- x5 `1 j2 L6 m(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
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& W9 g5 V5 L7 e# s' i  E# pFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
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' T( [8 Z7 A. ebC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s

3 I8 J6 m$ a/ q0 N8 ?
% M$ F: j! j6 ?9 Y& sintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
  Q2 b6 U' r$ [, |which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
; h) p- s( ~0 `' ~+ Gtherefore:

{(a+bx)/K} dY(x)/dx=Y(x)
2 l8 S8 ~1 \- I7 o
- C) V+ A4 X1 y( m: f2 R; Zfrom here, we can get:

- b! X" o2 \2 l: X$ ^- pdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
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so that:   ln Y(x) =( K/b) ln(a+bx)

( f3 ]6 a# U4 ~: U. Othis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
- n6 I% U9 W! v8 t( A2 k; K
1 t" R$ n9 ~$ i) W) s- {, }-(k+b)C(x)+s = (a+bx)^(k/b+1)

2 F0 x) p- S, s. i8 Z) Dfinally:
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1 r( Y" ?7 G9 |% v, L$ I% R( @C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)