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）

% V& F$ o2 q) ]- ~8 W: u# Ld [(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！

[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s

7 v9 j& V" t1 y* ]1 L! D(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
+ v% C* w7 d8 r2 @$ L. U; ^(a+bx+kx)c=sx
4 L& j" Q3 k+ l( T& b: Wc=sx/(a+bx+kx)

十年移民路_

Solution:
: ]  A$ O5 h9 w8 k/ J  B
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
1 C. c1 L/ N  N6 y; |so:

: e0 W  {- P  h9 B$ B7 PbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
$ y6 k$ d3 t! Q5 Ui.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
+ a0 F  `: t8 r/ W8 t
( q. g! @2 m$ |  |5 O4 |( Q' {
! S8 Z; W* ?4 ~5 @: X( Iintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
& y: Y6 q1 A; N3 wwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
( b& h7 o: i# F  V5 jtherefore:
1 e  R* q  E0 D$ T1 _2 z( I
5 n! x' ^5 T6 {' a1 k& G{(a+bx)/K} dY(x)/dx=Y(x)
/ ?, _/ Q- |8 I8 {4 w$ N, l
from here, we can get:
( S6 c; b2 K' q3 G6 w; M' V
8 I) P- P, X# w& T$ W, s* \/ O( `dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

5 P; r2 |- o, T' k, Sso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
4 d2 g  A2 }3 z% T8 a$ a7 hby using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

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