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）
- }3 j! H3 f  d5 T7 ]9 Y% P3 Q# M1 @
d [(a+bx)c ] /dx = -kc +s
4 o; u" Z/ E: K9 r( fwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
1 l) f6 N: v+ [+ j+ G' P
4 Z6 f8 z. G4 Q. J3 M* `$ Y多谢了！
6 r: a' ~, A6 y! G
[ 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
2 [2 o$ t+ V0 D0 i" _2 Sac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
5 b' |+ b% C' I3 |7 s3 L8 _8 k7 o
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
- V3 N* }9 c6 Z  d6 V: I  ^0 z
( i& [0 \3 Z; D  X( ?2 L) `9 D
) I9 m7 C; K% O6 ointroduce 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
- t+ T+ [& H% T9 `( ktherefore:
+ P  P. o. X  O  o" x& L, C
, z4 ^& k: h6 Y7 s' ]{(a+bx)/K} dY(x)/dx=Y(x)
: _2 s/ W, q+ S) s8 a% c
from here, we can get:
. ^9 {5 M. a/ x0 h3 j
7 B: I; c0 e, O' b, pdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

so that:   ln Y(x) =( K/b) ln(a+bx)

5 `9 y( }3 U) V' ]8 ethis means:  Y(x) = (a+bx)^(K/b)
5 w' I. Y# {# F# I3 s; Q) Lby using early transform, we can have:

/ @( Y0 i" }' H" T0 ?4 ]3 ^-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

* l' S& r1 Y4 i/ I! y- C- tC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)