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+ y. p' p+ c* g% W4 ld [(a+bx)c ] /dx = -kc +s
, C# F/ J1 R2 o# v' C( ~( y9 }where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

* i: N8 ^) B% z6 y, G. v多谢了！
% D& y; p1 B" L. p, q
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
6 [) ~6 }9 R% Y/ M. h4 K# S5 U
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
- {' x3 ]' S! H0 E(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
. v2 Y* s* X$ \
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
5 ?( t7 |7 c* U4 y' tso:
" c1 k+ ~" B/ K4 {) ?; |& d
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
  f  m# N: W; G" a9 ~: l
+ O* W* c$ E' l  d- A
/ A% Z  n, y4 G5 tintroduce 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
9 @1 Y3 I0 a0 v3 z) A: |therefore:
4 T7 n' `" w8 ?, f6 B( r$ L1 u
{(a+bx)/K} dY(x)/dx=Y(x)
9 s+ l; e* H) L6 P8 x1 K& h
from here, we can get:

+ N8 r! @+ [: L' s& `dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

0 |% S* Y5 P/ o8 l4 n5 Gso that:   ln Y(x) =( K/b) ln(a+bx)
: \2 S1 x5 m! s" Q3 e, F+ _2 ~% x
this means:  Y(x) = (a+bx)^(K/b)
3 Y5 z) W) d* s, |+ A( Vby using early transform, we can have:
$ U2 }! C2 J) Y9 j0 p1 k' Z. G
-(k+b)C(x)+s = (a+bx)^(k/b+1)

/ f' d; m- o- {2 X0 B/ Yfinally:

5 a# f( C; d  r' |- s" iC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)