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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2 k4 }6 d% M% R  J, |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?
! v. y  R& g, Q* R
多谢了！

6 }) `6 Y5 N/ r9 _0 D+ t$ s[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
9 h. x  Y5 _* h: s; [
; f8 L2 O5 k+ g5 Y% Y, C2 s) |(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
1 c! m2 O6 T1 gc=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
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bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
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(a+bx) dC(x)/dx  = -(k+b)C(x) +s
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) e, A* i- A3 r1 u. ^: kintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
0 e* u: E2 M9 Z4 ^) O  f) J& {& ]/ nwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
0 N  P# z. H9 O: u) ]
7 m5 l% l0 I1 K! Q: h- F2 Q9 ~- }{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:
; P1 `' K" T2 H( L$ h  |( D. j5 o
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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0 W; z, |' @; e- g% r& M: f1 F& Yso that:   ln Y(x) =( K/b) ln(a+bx)

2 H. s, D0 W  s6 T/ `9 qthis means:  Y(x) = (a+bx)^(K/b)
+ ~9 H! D% I# @. e$ Qby 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)