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）

6 l+ a3 ^: r# _- s' ed [(a+bx)c ] /dx = -kc +s
" u/ l. N5 i1 Owhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！
3 c* H! F& ]& ?" j$ B- P  t
6 j, S+ O; p; K2 `- z[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
6 v1 m5 ~+ _4 e( d; I( }) l
" h& E+ K1 v' \0 _2 I9 c7 P' R(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
0 o' B7 N4 g& ](a+bx+kx)c=sx
9 G: ]' z$ y% {( t3 D' j0 g1 ]c=sx/(a+bx+kx)

十年移民路_

Solution:
3 [0 t2 O# l9 S6 A( H% R
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
; G# ]( }3 Z) U( ^; j* qso:
: Y, J/ H( a( W
9 h4 w5 `; y$ C; O5 d- x! X# EbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
+ @+ q& }2 k4 @) ci.e.

" }; T" z) c, [8 _; I* i* [(a+bx) dC(x)/dx  = -(k+b)C(x) +s
; B7 I1 E+ t8 l6 N0 \) ]- H
2 q8 E, C$ C" R1 _6 p+ n
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
5 D$ ^" Z4 T: Rwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
' m9 M; y, |4 L  j- Atherefore:
* P3 z4 a# g7 d6 v0 ~9 t
# D0 O5 c0 G8 E! N2 z* [) O{(a+bx)/K} dY(x)/dx=Y(x)
; k) g0 E: i; d7 _: ~
from here, we can get:

8 K) G8 k. N9 v* D/ kdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

) }- B& }- H4 n( Zso that:   ln Y(x) =( K/b) ln(a+bx)
# T. ^9 {$ f  i* d; k  L
7 o1 o2 b6 @6 @, J# m- Q% F6 Xthis means:  Y(x) = (a+bx)^(K/b)
- S2 j& D' Y; u9 Kby using early transform, we can have:
' X- a- `+ l/ R. r0 `9 e" n4 R6 j
+ {  X( d7 K, c/ U9 I-(k+b)C(x)+s = (a+bx)^(k/b+1)
% b* T0 m- |+ d" s8 P
finally:

. M% Y, D: i" q' J/ z- X: e6 A$ j& hC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)