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
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

% f: N* O# s' T/ O4 |: H$ j$ l3 C多谢了！
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/ c& ?7 x0 ]2 G0 K- k7 n[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
' S) g3 A, l7 q3 L* h$ M6 _6 ^
* ?# r4 a  Y7 f/ U( }) a6 x(a+bx)c  = (-kc +s)x
8 Y  v' i) ]. j  Qac+bxc=-kxc+sx
(a+bx+kx)c=sx
0 G  d; I( W2 O7 `0 J# ]c=sx/(a+bx+kx)

十年移民路_

Solution:

  x6 x9 E6 Y  dFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
% S- S. j- Q/ ^' h+ i' rso:
8 Q' {) ?' I1 w- T) H7 `$ Z# E+ k
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
! r4 B: L) r0 mi.e.
& T: s" q0 Q( I8 w" k, I
6 m$ ^8 T8 g) @# _$ F3 ]& f(a+bx) dC(x)/dx  = -(k+b)C(x) +s


introduce 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
: N1 O: [7 r6 R2 \therefore:

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

from here, we can get:
' Y7 C$ k: P- Y# r  {* g% o
/ b* Z  r& s2 D) mdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
  [+ f- u5 \3 _' O
so that:   ln Y(x) =( K/b) ln(a+bx)

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)
$ {" _' ]5 Q" q) ~
/ C1 J( b4 J. E; u" J5 k8 a/ B# Lfinally:

- W" \, }7 [/ L* S' ZC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)