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）
# k% M; U1 n, G* S
* k8 p8 d. L, G9 M5 dd [(a+bx)c ] /dx = -kc +s
' r7 J3 c4 l- G/ t4 Ewhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
) k' A0 \0 n; U# Q( n8 ?
. a! S! _3 s# M- Q7 q多谢了！
& F* x7 A( k. ?" T* H; |
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
* ]' N4 G9 @6 J$ c; E- f7 h1 e
% V# ]! o: Y& E& \(a+bx)c  = (-kc +s)x
$ q! @7 T8 p: p2 ~( t! q0 Pac+bxc=-kxc+sx
8 {5 X, C% ^/ r9 u$ V(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
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3 A# x, _* n) g5 D5 xFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
5 Q" m( H/ z. v9 w3 ?# b2 g8 T5 zso:

( s. y' P2 Z; @; i( h* V1 v' lbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
( j7 {( {% e1 d9 a* s6 Y* {9 j

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:

{(a+bx)/K} dY(x)/dx=Y(x)
2 F0 ^* a0 @2 I  K2 b
from here, we can get:
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dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
/ Q: G$ A* y3 g- W
2 ~5 ^0 w6 m  P0 rso that:   ln Y(x) =( K/b) ln(a+bx)
3 ]. Z( B9 C& Q1 U  Z6 n
  M# m; q+ o1 F. S' b: \% [this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
8 _- `% x' b5 N6 P5 h
-(k+b)C(x)+s = (a+bx)^(k/b+1)
& U2 C7 F/ C2 {9 P" u2 l( z( N# n0 f
finally:

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