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）
  I1 g/ d' Z- Y/ b0 ]
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?

0 S# `% F' D) Z1 R- y; c0 u4 }' Y多谢了！

2 A3 y' w6 V; O- n; R[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
) c1 c0 T: }' W5 m0 i
(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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* v# E# H) w* f4 RFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

4 }3 p: D) n# x& i0 g8 FbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
/ _6 |% I) J3 G, ]: s5 T7 c4 `" Vi.e.
2 X. X6 R( e0 J  |6 P
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
5 ~) M2 i# S$ C) w. A& ?
: _4 G  h4 G) P/ }; r
; |% i% A; U2 iintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
7 ]7 P5 X- p& J9 Kwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
# Z# {* ]4 T6 x5 ]( H) H. g- Dtherefore:

{(a+bx)/K} dY(x)/dx=Y(x)

% }% ]- k" l( e% [0 z, ofrom here, we can get:
9 w# G6 C& s" t3 J8 P6 ]6 C
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
3 @2 h, g4 w0 i7 @; M2 @( b7 j
so that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
) H% r* }2 @4 \, N4 wby using early transform, we can have:
. A3 Y6 v" r4 s+ W2 j) W
-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
: E5 x: D! f8 p0 O0 i& M+ \
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)