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）
% q# ]% M+ j( ^2 W/ P4 M5 @! K4 \
- O- e0 y9 z* a( x. t. `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?

7 _6 ~( |/ [7 L! a2 B# P- V多谢了！
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% j0 `$ `4 K7 J+ i& A6 \# h  z[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

# @. [4 ?' ]4 ^: Y(a+bx)c  = (-kc +s)x
: R% Y* @! ?; a8 j& q5 M) @ac+bxc=-kxc+sx
(a+bx+kx)c=sx
; Q' C' B! c' q# jc=sx/(a+bx+kx)

十年移民路_

Solution:
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From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
+ v9 L# L- I1 v# vso:

  G  e, N/ T+ }7 WbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

; A' d; _) t/ r- ?0 u(a+bx) dC(x)/dx  = -(k+b)C(x) +s


introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
3 ]) \2 V6 w- z& Wwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

2 l! w; p# M. L' j; [{(a+bx)/K} dY(x)/dx=Y(x)

  J! w3 l; L. ^from here, we can get:
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# [% i, f6 k4 sdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

- N* s8 t" _- N/ @- ?; gso that:   ln Y(x) =( K/b) ln(a+bx)

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