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 Z, w) S% O- @8 ]
4 V3 X  W  C& S; B" n, e$ {' a多谢了！

[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
; y, `+ Z" h; n' ^' j
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
4 c  u* N7 K$ p(a+bx+kx)c=sx
# t* J2 i  w8 ~% a! ~3 \c=sx/(a+bx+kx)

十年移民路_

Solution:
8 \! ]- T) s1 ]+ c) `) b
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
; W) f- O' {, D; L" t' W9 `) qi.e.
! o( E1 ?1 \' e: y; v3 f
(a+bx) dC(x)/dx  = -(k+b)C(x) +s


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
3 e6 O* g$ A0 l  m% P2 r" etherefore:
% `5 t$ b$ @5 `' u! }: T
5 x, c3 k( G+ }6 t4 _3 \3 r{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:

1 `! s8 J% i- |8 G9 ~- EdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
8 ^/ ~- B) F+ o7 ]1 X! c
( U0 R& {1 B# w& u' x. Q, Fso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
  V- I, X+ o/ b0 v0 |" l; W8 T
/ F; Z4 N6 g; N7 v-(k+b)C(x)+s = (a+bx)^(k/b+1)
8 Z4 t( e, e' q7 B6 }! L9 }
6 s) v5 M" F; y% L! A9 M& sfinally:

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