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）

" B$ R$ i& [" ^1 u1 dd [(a+bx)c ] /dx = -kc +s
8 V) A, y% _$ b: rwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
1 @' y" Z7 a2 h' }6 e; \
多谢了！
4 l1 f* e" ^0 w6 v- Q4 t; s
; X0 Q! ?4 v" v) r[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

(a+bx)c  = (-kc +s)x
1 D/ s0 S3 l$ W9 V( ]* xac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

" ]# X5 [2 }9 G4 M% w5 gbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
4 @* f) q: g4 c9 T3 ei.e.
8 g* i& Z8 |3 m  A1 r+ N7 E
! ?" \9 R) L9 q. h(a+bx) dC(x)/dx  = -(k+b)C(x) +s


introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
: l! D) D* L/ }6 l. q- a7 N% rwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
  m# V" h+ s8 t2 ^% f% W" O$ E9 ^therefore:

{(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)
/ @5 }+ C- R( I  n* R. ^8 y
so that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
- ?; L; C& n: e  n: B- D5 x
9 ?  p5 K. h1 \- j4 Y4 \% f1 z! ?  V-(k+b)C(x)+s = (a+bx)^(k/b+1)
$ g  L& M$ N: u
# W3 k8 ~- ^& z4 h: p" V! Ofinally:

3 h* Z, A; y8 f3 j/ O1 k5 kC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)