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
, o1 Y% V, f7 Nwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

- \: j6 O% X5 B1 v1 D: O" j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
  w) d- i. Y4 @9 b# F  F& v. W
(a+bx)c  = (-kc +s)x
( d# h8 ~  T  eac+bxc=-kxc+sx
(a+bx+kx)c=sx
; s- R! q: O- o7 cc=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
% o  U  }, A6 ?3 |so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

# Z. ^- a+ W1 _( S: P! o(a+bx) dC(x)/dx  = -(k+b)C(x) +s
  B; [$ }* ^8 q1 r  u. r
# h  m8 P: a" N8 a0 [
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:
, Q. s  {5 N# `/ H' I, }* d$ v
{(a+bx)/K} dY(x)/dx=Y(x)
5 g& X+ v' E8 u. C* b
; @, {! `6 ~! ~5 Kfrom 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)

& O# \) c0 R& \7 u$ e/ sso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
! d+ R4 I1 E, z' C
+ A. R: E6 J1 {) c7 c# w-(k+b)C(x)+s = (a+bx)^(k/b+1)
: U; ]; t( }' D% A& s& D" I
finally:
2 _+ T6 N, [# \% N! o7 C* B
# Y8 ?$ H6 y9 O# s4 _8 nC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)