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）

d [(a+bx)c ] /dx = -kc +s
/ ]" {" f" G7 Q0 V$ o+ C% _; Rwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！
5 N& W0 B  F2 @0 ^0 q/ N; V1 H2 ]
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

& W7 N5 |3 v+ z(a+bx)c  = (-kc +s)x
, }. W+ }; s' A0 [$ Xac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
% `9 Q/ e# M1 U3 ^4 l+ ?4 r' r
0 X+ }7 e1 I. {From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
- s# H: `  `0 e  }' J  h4 [" q' ?7 ]; b
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
3 S7 ~. z6 V  _7 W) H( Z+ D2 s8 s$ oi.e.
7 ^" G8 H" V! {& ~
(a+bx) dC(x)/dx  = -(k+b)C(x) +s

1 `( H! N6 ?: O4 H$ d" f1 g
introduce 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:
0 H; |& G: R) q; E8 O4 i' o" i: }
0 [/ o3 f( N, v* A# I+ z3 M{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:

dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

8 [+ n( q! S" g& _( E; W% `; ]# W% kso that:   ln Y(x) =( K/b) ln(a+bx)

; {; H6 Q8 @2 o, }$ c  pthis means:  Y(x) = (a+bx)^(K/b)
8 y- e4 W8 y' w2 sby using early transform, we can have:
; j) N: P9 k0 M0 c) D; b  l
+ d- c  _+ v  l3 y" n' K-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
0 X) h# G7 f1 K
# O! T& V7 Q2 K' c" ~C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)