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）

3 y5 A. I+ t" `, v- Od [(a+bx)c ] /dx = -kc +s
9 h7 s! N, U* X) ]. K' C5 dwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
/ X- d# G- [* {! A1 R3 p
+ ?$ d, S8 K9 ^多谢了！

- A" S  D- M5 j" R; U; W5 e9 ?/ n[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
0 c- H. t2 W, j. K
- M) I: m& b* A(a+bx)c  = (-kc +s)x
# D/ C" {) R1 O# |2 Nac+bxc=-kxc+sx
(a+bx+kx)c=sx
3 C6 _( y: o9 y% s" h% A. v% vc=sx/(a+bx+kx)

十年移民路_

Solution:

2 Y5 a- s3 x6 m; j9 V! y% XFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s

# }) U( c( R2 A+ r% u  K" X& g
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
6 o$ _" N1 T) L" ~$ `- R0 }therefore:
% G3 R7 O  g- C/ ?* m9 I. S
! S) s2 w2 z: z! l7 w{(a+bx)/K} dY(x)/dx=Y(x)
. c0 A* \2 D- i3 ?, a3 ^3 o
& q7 G, K- ^7 l3 p6 |) Ffrom here, we can get:

dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
% f7 F2 c4 i2 H' o3 J% g  [
1 Q! C6 |2 C) t: Bso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
: p' G3 G- h2 Y  p9 A7 |: i
-(k+b)C(x)+s = (a+bx)^(k/b+1)
  b" o+ J8 H' w
finally:
8 L- [, a: n" l" e2 S
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)