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）
/ r( g- d, Z/ n! Q+ }  k
! r# S, J% f- f; nd [(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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, r7 E! p; Z& j' M多谢了！
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7 w9 n( u% b# U% c  J[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
  y* r9 o4 ]) [0 c
1 P9 P: f' d3 h9 e4 {* X(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
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( L) T! z6 I" n7 X4 p6 ZFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
! {0 Z9 O9 B- z: hso:
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/ N9 M4 ?, X( w& A5 P7 j3 t. {6 Z' VbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s


8 s) l' l- t" e3 s+ W5 O- Tintroduce 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)

. f0 ^8 a- Z( {from here, we can get:

" ]% D# a7 X# d! ]2 U/ n6 T% tdY(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)

this means:  Y(x) = (a+bx)^(K/b)
5 T; X" y( A* o7 t7 a, D) o6 `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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1 o6 j/ x. S% Mfinally:

5 Z( ]" L* w4 [8 [2 rC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)