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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1 g, v; m" B7 M' x) a9 ~+ Gd [(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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4 _$ }# {4 s) ^' |+ u; N[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s

' R$ ?, c4 P. m(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
* V- t9 [( `9 @- a! Oc=sx/(a+bx+kx)

十年移民路_

Solution:

2 u# [! h) S: u& e* d# L! `From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
+ h4 p2 e1 j' l( F# K% {so:
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bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
" K* m# {4 x( r4 ?% j: V- Hi.e.
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(a+bx) dC(x)/dx  = -(k+b)C(x) +s

& w3 F& B0 t$ _6 i# {
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
2 C8 R' p7 O+ X' T; k) Z; fwhich 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)
  O2 W, t. e' j7 P1 |- [1 j
from here, we can get:

$ P1 w6 i4 X1 ~. s: k# p, @; ?dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

. d! {' _, p4 R1 G  b" S+ f" F6 aso that:   ln Y(x) =( K/b) ln(a+bx)

: w" \8 n- |7 F6 X9 N3 Jthis means:  Y(x) = (a+bx)^(K/b)
7 d- _7 U1 v  r) v# W  F9 @( Zby using early transform, we can have:
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7 I1 w8 o& E: [-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
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$ t# x8 u: Q4 RC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)