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）
/ A; ?6 S9 e9 j) u! X
d [(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?

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

6 ?% Y# i" u( a(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

9 {, u& s5 e* V, U' {From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
4 K, q7 ?( U0 J; sso:
% m6 K* U/ C$ M2 k& p0 ~0 i
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
% U- ?3 ^( E( L4 `2 ~- _. Q) r1 fi.e.
, o, l) U! S: b2 A2 v
, V* A, ?$ P6 ~: s(a+bx) dC(x)/dx  = -(k+b)C(x) +s


introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
4 W- ]8 N- _4 u$ Rwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
0 F$ ]$ u: E, u: @% Y1 dtherefore:
7 G4 B+ [2 k# Q& I) v2 T9 w
8 C! P% U! S7 R{(a+bx)/K} dY(x)/dx=Y(x)
" Z7 v$ E, h" l7 p( 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)
2 ^0 t+ I8 M8 v- s; x. O2 h5 `
so that:   ln Y(x) =( K/b) ln(a+bx)
# k$ S. Z/ ]: T  D9 C* P* `8 z
6 [5 e' E4 x+ I5 U+ \this means:  Y(x) = (a+bx)^(K/b)
3 f  P( T) }1 v% P/ P% o7 `( Yby using early transform, we can have:
: F6 o8 U6 A- }
: F8 Y& u# [* S; Z. ^-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

% ~- N% ~8 b! \+ S  e; G# NC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)