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' m/ U; F8 m, x+ A, Y1 ~d [(a+bx)c ] /dx = -kc +s
* J$ n6 V6 k! P4 [) s, Gwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

2 U- z# O  t  \多谢了！
3 f$ f6 A' x2 B5 |7 B& |
[ 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
ac+bxc=-kxc+sx
) M  y$ _# a( Z9 B5 x  P(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

( w' a) L: @# ?7 w5 i# wFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
+ i8 O8 x+ y% w  @so:

- f( c, z. w9 L5 C/ g/ qbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
5 _2 \6 @2 y7 u' P' O; b# R" R
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
# L4 A+ `: p0 b7 ~+ q2 q) W

" v" g; w( i- G) D* I" Gintroduce 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
7 b! g, h- A$ i3 v( gtherefore:

, j- \6 Q2 h3 ~  R( x& n. E{(a+bx)/K} dY(x)/dx=Y(x)
& X/ h! m9 Z, A
, t2 b2 u1 w6 W: A0 A7 U1 Jfrom here, we can get:
. X( @& D  E3 ~1 j+ i: z3 K
# {1 K4 D% N% F" ]8 J# cdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

) L; n% S. f" D9 Aso that:   ln Y(x) =( K/b) ln(a+bx)

% W2 m- l! j+ s3 o, Qthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

& v3 E* n7 Q: c2 Q) n-(k+b)C(x)+s = (a+bx)^(k/b+1)

9 x5 i) N# q( T( K# Q5 U( \  @finally:
9 U; J5 B, C  C$ M9 {
; q- U+ v# A( G7 r: H; d. j' lC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)