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）

# j  W% y, I* a9 C+ `- gd [(a+bx)c ] /dx = -kc +s
4 o1 _7 `+ X# {6 f0 twhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

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

(a+bx)c  = (-kc +s)x
$ K+ Z; ^8 H: z& z9 m9 m0 Sac+bxc=-kxc+sx
# f6 W: ?- e% \5 X(a+bx+kx)c=sx
5 X% Q, |" E6 c" o2 Z* ic=sx/(a+bx+kx)

十年移民路_

Solution:
* q6 y7 v9 P, _' i( J4 H
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
  G; a* g+ l1 T/ p4 e
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
: s6 W, p4 [% l! q; p9 ^
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
- @2 _7 k% y! `" x
# }) e0 D7 V& o
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
6 @. D0 q3 m- d& y" D7 N1 {which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
6 l+ x0 E) h, ]" m+ h( S, L7 A1 btherefore:

{(a+bx)/K} dY(x)/dx=Y(x)

3 h  l0 F' e/ pfrom here, we can get:
* h2 c) \/ Z) _" G5 z
# d, V8 Z  N$ V* D$ P) J3 V* NdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

6 U# x: Z- l! M/ }, g% f, W+ ?so that:   ln Y(x) =( K/b) ln(a+bx)
' a" }; a2 a0 V; w  l
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
: o9 _2 H4 l* u  x8 z1 H
-(k+b)C(x)+s = (a+bx)^(k/b+1)

6 [) p/ y5 ~' y& {finally:

C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)