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?
0 e' c! A: ~) X
多谢了！
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[ 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
& W# v! G2 L3 t! M( A. sac+bxc=-kxc+sx
4 S' T! |4 ]' t) ~( R, Y& p7 x(a+bx+kx)c=sx
' M. U7 K2 P* r3 g& N0 yc=sx/(a+bx+kx)

十年移民路_

Solution:
7 K- R) n# r! l8 a
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

0 _$ o8 z- {6 J$ obC(x) + (a+bx) dC(x)/dx = -kC(x) +s
( V; `3 ?: X, T+ X( O7 N1 j" \i.e.
+ A. f* q! |) L8 t# e4 ~7 S
+ X7 G3 h: x0 E0 p8 h  Q(a+bx) dC(x)/dx  = -(k+b)C(x) +s
6 @& l! L% ]( L6 ~3 E! r3 v' F7 ]

7 D3 z- y3 {" m% Eintroduce 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
9 |* t$ h4 x! [- F7 B' g: A7 V3 Ftherefore:
" n6 r& y) U( m$ S
' `3 Y# X1 Q1 j3 r{(a+bx)/K} dY(x)/dx=Y(x)

! x- C6 l" C! H2 b/ {' v  efrom here, we can get:
  Y5 @5 v: J9 i# W
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
- Z. B7 ?4 X7 w2 G. r
8 V3 V1 I' {: U- Q( t7 w! aso that:   ln Y(x) =( K/b) ln(a+bx)
: E! g' y, S1 s. U/ w# A5 o- r! D& p. B
3 |8 }6 ^0 x1 B: q# F2 h% Hthis means:  Y(x) = (a+bx)^(K/b)
5 B8 ?& E4 ~: f& \; M) ~( Z& Pby using early transform, we can have:
; }$ O6 H4 T8 W& p! H
-(k+b)C(x)+s = (a+bx)^(k/b+1)
$ x) A! ]% S6 n4 B* D
3 m) \9 c1 B& t, S' Wfinally:
; a1 p( s9 x4 l; h' a: H
. f) p- i: a$ u. j- \) JC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)