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?
6 K3 T& u  @. ~; Q/ P
多谢了！

% D7 w4 d6 S! {2 A( D5 b[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

+ I; Z" @3 g7 U, v& D/ O; B9 v(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
( l- I8 w5 L$ B0 G(a+bx+kx)c=sx
% V* z+ b- n9 v; u' G' E1 Q' ]c=sx/(a+bx+kx)

十年移民路_

Solution:
$ \; l+ P1 u8 Z# `/ u$ A3 X
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
4 Y6 V+ {* ?/ `5 K" m8 lso:

. h/ Y$ i+ ^3 a6 R' B; [' M2 hbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

' v7 ^3 ?+ g4 \- z* n(a+bx) dC(x)/dx  = -(k+b)C(x) +s

' o+ R7 ]# Y# M7 H6 ~: O
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)
6 O" @1 x' f% y9 G; U
from here, we can get:
2 S0 t+ B( K( y9 K: Z) [
- n/ M# o1 x: v4 r9 f' \dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

so that:   ln Y(x) =( K/b) ln(a+bx)
6 I% I" _! v* ?! \- c2 ]3 _" j
+ a. J! L) P, V! p- j; K& F. F( Xthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
( \. @" h" Z$ j  g- ?$ e" [. g
-(k+b)C(x)+s = (a+bx)^(k/b+1)

9 f8 y: s% M0 H/ H9 J# cfinally:
1 T; Q3 O8 E* s3 i( H5 w6 x" Q
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)