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
% O# M0 J- ]/ `9 p/ Gwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

1 ^( ^' X, a1 ]9 r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
; Z& s7 f4 }( W) B0 p
: O8 o) c) p3 }$ d* k(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
0 J; C) D# N9 S& _! W: K) E(a+bx+kx)c=sx
6 j/ c$ m) h% |7 ^# Ic=sx/(a+bx+kx)

十年移民路_

Solution:
0 \# m. [7 c2 e5 s$ m* {6 a. K
3 T; N1 j, u7 [; X1 F2 r( WFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
, E2 U4 q, ]/ |2 ?& ]; E/ J# Kso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
6 y. p$ N( D* f! d% yi.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s


& k. E( K) i" G2 ~2 Q' c4 Iintroduce 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:

# {$ n) u5 l; T$ y{(a+bx)/K} dY(x)/dx=Y(x)
9 l  R8 g7 K+ y8 Q- _$ M
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)
' f- ~4 \+ L: I6 m2 U4 B/ Y' s
! m4 h- E/ j# L* Xso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
2 p  ?) y( w, z  w
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)