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）

( o, a! g% G9 g" }, Kd [(a+bx)c ] /dx = -kc +s
$ N  c( E5 }- C, J& Mwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！
( Z0 y/ K3 m& w) W% ^9 @
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

8 K8 `4 J2 o2 a; g) t* _8 @( k(a+bx)c  = (-kc +s)x
/ h# C& f+ i! z9 p- Z: U# V3 Wac+bxc=-kxc+sx
9 s; ]+ t  P; p1 s(a+bx+kx)c=sx
2 }0 S/ `! K" C( C" n# z) Cc=sx/(a+bx+kx)

十年移民路_

Solution:
* t2 J1 G: z0 k
' A: {! n$ ~8 {  T: oFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
5 S' \7 Q: h) a, i$ C8 Iso:
% j# k' t3 T! P. `) b
" j' ^1 Q* E, L+ q8 }/ Z" |bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
0 Y9 r  `' m( Q* r' \& Z9 i% pi.e.
1 M  _) F( f% I
$ n/ K$ }8 F* A9 Y+ k(a+bx) dC(x)/dx  = -(k+b)C(x) +s

/ {' Y2 G7 J5 q+ k
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
- ]; O( {  N4 f6 X* t) r- Z8 F0 C4 |which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
; c! C0 h; w% ]% _- l# ~therefore:

; L3 z7 U9 a" H0 H2 B7 o{(a+bx)/K} dY(x)/dx=Y(x)
6 p, a5 R' ?9 G1 N3 |
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)

( \: A/ C! S) Y$ D; P4 Pso that:   ln Y(x) =( K/b) ln(a+bx)
; ~3 _1 H/ w3 T$ i* f0 O
& w8 d/ P6 D0 cthis means:  Y(x) = (a+bx)^(K/b)
# h5 t; L; l$ Z5 S  t' x0 b7 uby using early transform, we can have:
8 h" C& H' s) {( r1 b
-(k+b)C(x)+s = (a+bx)^(k/b+1)
  q: `% J$ `0 M! u( H- H
finally:
, z% q, o6 l* Z3 q4 V
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)