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）
5 _/ S! Z2 B) y5 L. [% R. q' h8 L" J" w/ @
( [6 `- }2 N  Y% c% @1 Zd [(a+bx)c ] /dx = -kc +s
: f0 i) A' |# Zwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！

, |2 j- u( W& X1 z  @[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

5 |4 b- i4 ^* y6 {# @, v7 e: `(a+bx)c  = (-kc +s)x
( X  o" E" h$ Eac+bxc=-kxc+sx
, Z! {: d: y. w(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

4 c7 D2 D4 v8 f' p* K( Y5 b2 qFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
' p4 r/ m8 u( [4 t/ p- |so:

. ~' e: d( _( T8 a/ r( z- mbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
) N, @1 I$ i3 [( r
(a+bx) dC(x)/dx  = -(k+b)C(x) +s


* n- [2 F6 {4 o: p+ h! cintroduce 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
2 V9 @2 T+ c) H( X* ~- xtherefore:
! X/ Y/ v/ Z& p8 L  C+ x0 g( ^/ y8 Z
1 g' Y" F  R, r5 Z# w  [$ s9 r5 s; n{(a+bx)/K} dY(x)/dx=Y(x)

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)
1 D- V3 i- F7 ?3 d
( P+ N5 z! h# a) xso that:   ln Y(x) =( K/b) ln(a+bx)
" M+ _" f: S$ Z. X* e  ^5 J
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
( W7 q8 R  ]8 X" f% N/ [' A
# w& S, ?3 l) i' c1 f  ~7 Z-(k+b)C(x)+s = (a+bx)^(k/b+1)
5 X& {( n  w4 n1 G+ J6 I
finally:

C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)