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）
( x6 @0 {7 I+ q) g9 `9 X2 j
! f" P/ G+ u7 t" U5 X( nd [(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?

多谢了！
' x' P* T( @) x. H* P
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
: F5 f' z+ l, N. p4 ^
9 z* g) x( x7 T6 x3 x5 j# V(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
! |& r& L7 y( p
7 h( ]. T5 t+ R& BFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

/ m4 F6 y) |5 y) _0 a- F: `% zbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
$ v0 o3 j" @. a2 \' l

& x" s% g, |4 Bintroduce 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
8 G2 O- ]. z" ?( k( n- E& wtherefore:
8 y+ ^4 x, N1 _* }' i2 j
  Q. V4 ~1 r2 d8 Z) u0 P. c* F{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:

dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
; [" l  [8 G2 l, _
+ S7 y5 U+ ^1 x' h* y" oso that:   ln Y(x) =( K/b) ln(a+bx)

8 z' z7 D# U5 |4 j6 ~this means:  Y(x) = (a+bx)^(K/b)
7 j2 j- V  o- ~4 q2 Rby using early transform, we can have:
& p+ x% D7 |3 K. ^3 [
-(k+b)C(x)+s = (a+bx)^(k/b+1)
! K- p' b* ]  s# h* |: s
# J9 F& v  f3 ~' c, Ufinally:
# ^4 y2 c1 @5 ^1 W/ m8 a( c
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)