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
/ R$ W6 a& M9 O! o: U* t5 f2 Kwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
0 i7 D4 d& I! H
多谢了！

; v, V; Y5 Y& z  I0 a3 Z6 i, S5 i[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
1 a( H$ R2 s  j6 }
% a; u4 u# W3 |' I1 i5 f(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
$ L; R, ?" Q6 y9 ](a+bx+kx)c=sx
& ~/ G& B9 Q0 Y0 kc=sx/(a+bx+kx)

十年移民路_

Solution:

3 P. Z; `2 P5 q" r! R; ~From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
' ]  p8 ?0 p" {+ E2 lso:
7 ?' }* Q) e7 d
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
$ E" M% a9 p/ p
# q, g0 X8 G# z# a: Q, C* t(a+bx) dC(x)/dx  = -(k+b)C(x) +s


+ W$ R$ ^) l& s3 o! k- ~introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
/ s% Y- R! S* a6 ?* wwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
1 A, X' l" H4 A& k% j( Ntherefore:

{(a+bx)/K} dY(x)/dx=Y(x)
# ?' q% K/ u( j
from here, we can get:
+ b! f# D) t. {+ [2 |  y. v( s( x
* M% J5 S# q! l  H" B. OdY(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)
4 q, V( F! l! P
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
% T; U7 u; M7 m$ q3 V. i
-(k+b)C(x)+s = (a+bx)^(k/b+1)
/ i" N3 A' C% \! U- `4 [" V
9 Y- i0 K1 D' Z  ^  y5 C' y+ @finally:
% |, ^0 D7 [4 ?
* C8 a! E" ?3 n) K. r) fC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)