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）
3 `% C( O, A/ W1 y0 Q; Q
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?

多谢了！
" T* n- I. y( ~, \
: ^# S' b' w1 ?2 Y* I! k' Q[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
7 a7 Z# t# G3 f' @" b8 j% R
(a+bx)c  = (-kc +s)x
( s" d  N& \( O3 nac+bxc=-kxc+sx
( Q' V2 f) e# d/ k(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

( h. B' s. z: F; AFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
4 i  D4 W$ L; n: g( M" Uso:
+ [5 \/ n: V2 Y* D$ J3 Q/ c+ d6 y
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
/ F/ ]. {6 Q4 J& q  Fi.e.
3 W, h, n% q, N' Q1 b
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
7 O8 c4 S, H  g

* @- z  F2 b3 G/ G8 wintroduce 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
1 s& T8 M- W7 Xtherefore:

{(a+bx)/K} dY(x)/dx=Y(x)
& S( M/ n' B8 {5 S
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)
% v2 S4 |  ~4 B! \: B" S' d1 _+ o
so that:   ln Y(x) =( K/b) ln(a+bx)
$ i( t3 h* {, O5 ?/ J* O; U: W
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

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

finally:
" o, T1 d0 I, J* e" W7 B5 }* g
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)