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）

8 K# G/ [* A3 y1 K5 a4 `d [(a+bx)c ] /dx = -kc +s
( T0 [" o. v7 {9 Dwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
$ }9 y/ g9 d& L& p0 H
, o6 a5 J' [/ z) O6 H  P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
9 x) W, o. b! A6 |- I
(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:

& ^$ m5 v5 f; ]4 PFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
3 M( z/ u: a7 q! nso:
* w; y; b/ T! `2 r/ i+ d
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
1 F1 U+ J6 D4 p1 j( e$ hi.e.
5 f9 w& V; A- G: m8 {* @  G
4 M+ R+ l/ e& [7 w& _; L(a+bx) dC(x)/dx  = -(k+b)C(x) +s


introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
( h* m* R2 l# i6 \3 ^  Q: Cwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
5 J; ~! x  `. O7 @8 ftherefore:

{(a+bx)/K} dY(x)/dx=Y(x)
9 Z3 w, r( ~6 x7 U  N' v
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)
! u4 \6 E( r( C2 e& e; {) Q
* O# S- {! E, C+ b3 eso that:   ln Y(x) =( K/b) ln(a+bx)

& J7 B, j6 X: g& Jthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
8 x! L9 |8 B& r! p- s
-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
9 p9 L3 O. c  K0 T7 `' ]" C
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)