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）

0 B, h2 `/ G4 x% B  l. n6 _d [(a+bx)c ] /dx = -kc +s
! f2 U0 b. s& Pwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
5 h& E/ `/ ]  U7 {, J1 ?9 ?6 b
$ R+ v, s7 ~/ [$ x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
$ W& r6 Q+ V9 X6 y* l+ v+ ?8 V
" l- U# x/ _. h8 y6 r- S(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
# c* A2 l8 P" @) hc=sx/(a+bx+kx)

十年移民路_

Solution:
$ O; w& Z. c' h
0 `4 w( O+ Y* K, h; [: Q( ?From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
: }+ A; ]2 P8 z6 X" Pso:
2 u9 Z3 E) }/ ~# l  I+ `: P
& w& Y0 _# Q! c0 k- \- T1 R& Z5 BbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
( R8 e' w+ T1 Z( w/ C
+ a) \2 b3 L" l* b! |  v6 o- b7 W9 W
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
& B( R! |$ {0 {+ ]therefore:

{(a+bx)/K} dY(x)/dx=Y(x)
9 t1 S9 r1 `. Y: S1 u8 y
* b1 ^1 Z5 k  t" Gfrom here, we can get:
8 s' A  N- w5 k7 _0 T! f6 x
" p6 a, S0 ?( ^7 adY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
, x- P( G" P5 l
3 }: {" F4 n/ ~0 o% Dso that:   ln Y(x) =( K/b) ln(a+bx)

6 n+ c3 h" r* R0 V1 t" mthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
& {* B# I) o& j5 J( w* e
-(k+b)C(x)+s = (a+bx)^(k/b+1)

9 p- k2 ]  o8 V: Ofinally:

1 `# c% c# w& l1 D: _2 i& iC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)