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 E; H% @  p* s
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?
, l  ~2 e7 |; F4 M
多谢了！

* e0 {2 ^% N# Q' |& Y, n! t3 P+ ~[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

+ D4 @( O/ R- Y(a+bx)c  = (-kc +s)x
. v7 }2 k7 p0 y& Cac+bxc=-kxc+sx
& e9 `4 i- [4 F% g1 o$ Y(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
: c  k" v6 B$ A- M( v; P' r
! W% D7 U4 j7 b/ NFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
7 u% i$ V9 P, G9 ^1 d# p
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s

' u' o# h; }! W  t& Q
9 V& {; P& k5 M/ gintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
* Q/ d; d' E2 J* k- t! R( Z0 |which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
( o6 @4 U% \6 a, b5 J. a$ vtherefore:

: v. |, a% c* x, B- Y5 U" d{(a+bx)/K} dY(x)/dx=Y(x)
* O: W5 w/ q5 {( y; O( o$ D
2 m5 {# y2 D* P8 ?& pfrom here, we can get:

* o9 ~$ j* I* Z, ?dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

so that:   ln Y(x) =( K/b) ln(a+bx)
9 l/ c/ R" z1 x; ?, a
+ F5 N! {- s+ \5 jthis means:  Y(x) = (a+bx)^(K/b)
" V% T/ F& w1 Y8 J1 ]: y, ^by using early transform, we can have:

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

( Q( Y/ B( ^8 L& C6 i0 L: lfinally:

5 [& U0 o& S- |9 LC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)