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
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！
( T+ s& o! p0 H) O: f5 U( Y
1 J( P* j2 @6 R/ [[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
- n) r( o  n- I9 E
6 N- Q% Y* g! g% c. \(a+bx)c  = (-kc +s)x
; K7 G" }9 i& E9 Y6 h/ Bac+bxc=-kxc+sx
) s7 z0 @1 \8 [: N(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
! M* D/ L$ O- e4 r1 J3 T, n. @3 J
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
9 ]  q# D9 ?7 @" E6 c1 p, q$ }. ~i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s


introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
3 d" u8 h0 r- \( Ywhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
9 g# [* o" F  J' d. vtherefore:
  e9 l# Y' L; T4 s  V$ I' F
0 h1 M; F0 C, l' ^# g) E1 a{(a+bx)/K} dY(x)/dx=Y(x)

from 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)
1 O( f3 w$ q) P) ]7 B, K( o5 O
so that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
9 w8 q3 u+ j7 K/ o- oby using early transform, we can have:
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- m& D- M" `2 k* O1 O-(k+b)C(x)+s = (a+bx)^(k/b+1)
* S6 |) Q- p0 d: e4 u- G
: P% r2 m0 P7 V  F' e) Xfinally:
5 q/ S/ o  O* i# a2 d! ]
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)