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
0 X* i9 R; H0 \1 R8 X( ~where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！

# J% v9 ]& C& N8 ^% l- [[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
, U; g/ S/ h+ }$ \0 S6 ]
' X; U$ P% L% |) J2 L$ Q(a+bx)c  = (-kc +s)x
+ x  b" Z* t! a0 V' oac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
! P9 b" |$ ]- r" r: x7 C! D
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

% ]. q' m! X$ @' E/ ?: _bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
6 l3 G4 ~6 Q7 X; }- z1 V' Fi.e.

(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
therefore:

, j" e( p  `: t; l{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:
5 N; G! O4 ^  M$ ~# i/ x* ^
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

. I3 o" a5 u# g# `so that:   ln Y(x) =( K/b) ln(a+bx)
) I% y4 l7 `5 f$ r, m  t" x
% p" {: t' `6 l, D$ b9 ?4 y% Jthis means:  Y(x) = (a+bx)^(K/b)
5 i' k( @( S5 Gby using early transform, we can have:

& R' }6 E# D( z9 b2 v6 S-(k+b)C(x)+s = (a+bx)^(k/b+1)

9 }* Y- q6 }/ x; H7 W! L$ zfinally:

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