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）
% C  n' c5 X5 N7 y- c* j
- |% Y* k8 J$ i7 y  q1 T% f+ `. }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?

% Z5 s, V( i  W7 Q$ t" [7 f4 O% }多谢了！
$ h/ f& _" H! J, s1 u
7 k6 C# N/ L4 j4 p- w. _[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
( Q$ q8 p/ ?' r: h
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
" C  z* d4 K2 R7 w8 K(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
* d( ?# g) `2 [# j# Lso:

6 R  l$ }- Q) O8 R# f3 \7 t# dbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
/ h% Z( @' {4 E) L7 n' \0 B

( W8 }, m. w/ Z" ?/ ~" ^introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
6 h+ \. A9 d# I3 f4 T6 `. qwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
2 ~; I' g& a  ]% n2 {* {. S9 H/ x0 ^
{(a+bx)/K} dY(x)/dx=Y(x)
+ U1 _  E  E2 z* Z2 X% i
from here, we can get:
" G3 ^# [, y# g& `, p4 K
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

7 S* g: E' R7 o: ^: g; ^" rso that:   ln Y(x) =( K/b) ln(a+bx)
5 [/ ?5 \$ ?. ]5 e# f
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:

% z* S+ n! `( T6 B4 ^5 XC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)