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）

d [(a+bx)c ] /dx = -kc +s
: r9 p& S& ?* nwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

' V. h! k7 Q" L1 d/ A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
- r  Y9 A) }+ a0 Q
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
/ S6 [4 }) V& v5 R" O, q7 rc=sx/(a+bx+kx)

十年移民路_

Solution:
4 f; S# _! Y& ~- [
, C& E9 B- h3 ?From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
: F2 O4 E( z+ E/ t7 e- {
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
% s* C0 }# @* _$ C$ D
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
$ J6 P$ Y  G3 x/ }0 V& U
* U$ R3 @2 b2 y5 c0 l9 J
3 a8 c/ }5 M# F9 dintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
) }$ [' L) _% h3 V- [- rwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
8 @# S; t8 i7 A3 V1 H) Y9 Ltherefore:

{(a+bx)/K} dY(x)/dx=Y(x)
* ]' p# @! C. H" f8 N+ K1 P5 {
from here, we can get:
& _( S( T! v5 @. r7 T: F0 s% A
dY(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)
  e9 ]0 E: s! Q! N' u
this means:  Y(x) = (a+bx)^(K/b)
. d5 k3 W' y$ e" ?! M% I+ L6 Bby using early transform, we can have:

  v5 B0 K% w" J! C  ^. h-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
4 `! y$ ^8 w4 A3 R0 }
9 {8 W( T# S+ f3 ^2 |C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)