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?
1 X) d# }9 ~( l; D
多谢了！

. e# a( `7 h! J; t8 m: @* Q[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
/ h" p8 d% E8 L' c. z: J
(a+bx)c  = (-kc +s)x
* d6 W2 v" v$ E( l( {. Yac+bxc=-kxc+sx
1 r' a4 m/ ?( q(a+bx+kx)c=sx
9 c  ]! r8 [' f1 n; ?8 Hc=sx/(a+bx+kx)

十年移民路_

Solution:
% R8 D! @0 z, W  N
  X. G$ k5 P, U+ A  t9 f6 VFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
% P0 o) q) V- @8 J
1 o" U/ X+ |' u" e0 b5 e2 n0 mbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
4 ?: [6 S* {% R4 }7 G
! O! M1 Q4 Y# {1 a/ i" X6 Y. b(a+bx) dC(x)/dx  = -(k+b)C(x) +s

+ |: g. ~+ Y; c& p* N
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
therefore:

& U+ T1 K4 w: O, l( u{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:

9 m: W9 {) H" U" w7 O5 @dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
: G1 q6 T& O9 {+ I, ?& d$ r4 \! T* n
# ]# G+ m5 q" W3 f% A5 L0 P, m9 Oso that:   ln Y(x) =( K/b) ln(a+bx)
" O  j! n5 ^8 j. F: j: S3 @
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
1 V% t3 M& L/ [" Z; v
-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
  H4 m8 V' U% R" H
5 A. [( d" {& p0 J# ~4 d# J8 hC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)