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
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

7 j+ z& T( r0 A) s% x- Z多谢了！

) ]8 Q# h8 W/ ^2 V[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

(a+bx)c  = (-kc +s)x
5 e4 }# u; X6 [3 Vac+bxc=-kxc+sx
" x; @! ~2 ~! l: g(a+bx+kx)c=sx
8 k1 i* L! [6 v6 S3 Ic=sx/(a+bx+kx)

十年移民路_

Solution:

, e3 C) X" r$ ^2 e9 _6 B7 k5 I; D* n9 yFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
  {- E6 ^0 p9 l' c6 J
5 ~8 D- w7 z, H9 E3 S0 EbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
0 I3 R8 A; e( g/ K3 u& Ui.e.

" J; F- o9 q- k" `(a+bx) dC(x)/dx  = -(k+b)C(x) +s
1 E5 q5 z) D6 t5 f

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:
. z! ?; |4 a, C$ c4 T9 C* w
{(a+bx)/K} dY(x)/dx=Y(x)
" P8 N6 M" G( T
from here, we can get:

1 @0 C! ~5 d8 t  HdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
7 n/ m- p% i. `$ f0 G8 q( d3 f
1 v  H% A( o- w/ I- ^so that:   ln Y(x) =( K/b) ln(a+bx)
- o8 {$ C) g" g6 i4 O+ q
% m# w, w3 L  k9 V# H! lthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)

" X- j( {$ c, l% q4 }$ ?# r. zfinally:

0 n0 n0 s" n; q' W/ AC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)