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
: I& B8 N, b# c& f" Y) J2 V/ zwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

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/ _: A4 ?+ K% Y/ z
) k7 i2 v- l' i+ d8 L(a+bx)c  = (-kc +s)x
" ^+ S! H5 n$ [. zac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
* o0 R8 S9 q+ m+ ~6 f4 `0 e$ ~! n
1 S6 S$ d. s1 ]* Z/ b9 CFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
5 \& i3 G2 A* Z5 k6 Q3 K9 |. }$ [' X3 z
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
0 U% J+ p0 b" J" v' [

2 h5 j; c- B4 f1 C4 c; vintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
: c* g  C/ P; A4 P& k2 _which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
- O* d( b2 {4 p1 P3 vtherefore:
% b: s1 E+ n6 t* u2 J, t
{(a+bx)/K} dY(x)/dx=Y(x)
! m/ N. Y& l) U" n3 t) `, ~1 ]
from here, we can get:

- ^$ A( A0 H4 }6 Y# FdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
8 r- [1 A! `9 K! ]
so that:   ln Y(x) =( K/b) ln(a+bx)

. l9 y6 [- k' C4 S* }0 G  lthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
" L/ `% ^: A" S$ _1 n% V
9 ]4 V3 W* ^6 R% B9 a-(k+b)C(x)+s = (a+bx)^(k/b+1)

% m. f6 E8 x! J# x6 Nfinally:
; {5 {9 E9 m( P4 v, p1 p
7 C: H" O7 L- }9 e6 OC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)