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）
% j( F% Z7 m2 p3 h6 t4 \
0 J$ ~. r7 q$ B9 y' n  X" Ld [(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?

6 O3 Y3 y' B+ `. y" c3 C+ X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
" w8 Y2 }' ?* Q' C
. ?* I' i- _$ c+ P8 m2 A. `(a+bx)c  = (-kc +s)x
. p7 a( T# D3 n, F6 Qac+bxc=-kxc+sx
(a+bx+kx)c=sx
; O- ^3 {$ J, l* `; ]c=sx/(a+bx+kx)

十年移民路_

Solution:
* {' p, J$ V' l  ]% ]+ A
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
6 {' K$ h! J, i, M4 H9 ?+ f  _so:
* [7 l) e/ z. E; T$ L6 i7 {+ [
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
" }% y/ G9 ~2 Ui.e.

! X% F9 y7 E- a(a+bx) dC(x)/dx  = -(k+b)C(x) +s
. B$ ?  Z- i1 g' g, N% A  Q7 S

' E! V# _- F# c. `5 R! sintroduce 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:

{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:
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6 Q% n/ C" H4 i0 B: [0 g# e% pdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

# B! H% n: @5 ?- G' |! G9 e7 Yso that:   ln Y(x) =( K/b) ln(a+bx)
7 h) V; O9 K* q
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
) \9 P) B6 s9 b" |; q: [
& f, ?, o5 J" g) a-(k+b)C(x)+s = (a+bx)^(k/b+1)

' u1 W/ }2 R3 c  s5 efinally:

% I6 u5 O8 T8 c/ d. N# Q$ Z( O2 T  fC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)