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）
9 Q1 {! T# [/ a/ T% P( F) j
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?

/ L# }. ]& t5 k% q0 z" j# V# `/ H多谢了！

, x: Z, r0 t1 l, R8 A/ E[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

$ v9 B$ b5 x/ c+ E4 p% B7 W, t7 m(a+bx)c  = (-kc +s)x
5 z5 P0 K8 a( K) g& l  L# {- Bac+bxc=-kxc+sx
# t1 u* [+ w- \# h# K' I(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
0 ?1 a; @2 d+ Q' H9 w4 x' k
6 t6 S. [! x7 \From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
# `; G- Q) I9 W, Zso:
9 a. R( P) M0 Y" n1 \5 p
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
% L1 a# N  t% Q( k; i. [: Ei.e.

. h+ Y' k+ Z% d  m1 m(a+bx) dC(x)/dx  = -(k+b)C(x) +s
+ A+ _  H4 p- o& V. u6 Q
" h  O, |7 o6 g3 E* g
' y' R3 K  _# K( ^: j  ointroduce 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
* y: n- ]  p- t8 a' Ctherefore:
$ d; z5 ^# k3 a& P
6 `7 K$ {  c& e' y, e{(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)

4 F' q0 f% M0 }- ^. P( xso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

  y8 K5 o3 H* H! M  v-(k+b)C(x)+s = (a+bx)^(k/b+1)
" H# f: b& @$ w! j3 G( c( y
finally:
/ ?* n+ n( A% L  m1 l' p, I
, x" S( I( v6 AC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)