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）
* S3 K2 |/ O7 U4 \6 L
d [(a+bx)c ] /dx = -kc +s
: R) o" ]0 \1 owhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

1 i5 `) \8 T1 ^! `6 ]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
* @& [$ M- t6 Q' I# p2 n* g) P
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
5 r$ L& g5 k. |9 Q4 p# I7 x' ~/ g4 f(a+bx+kx)c=sx
" [  P3 e5 d4 t" F+ i! Oc=sx/(a+bx+kx)

十年移民路_

Solution:

( U& u% b$ F  v( a6 e6 b) EFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

4 o' a1 L  H4 `- ObC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
) f* M: p3 r( g. d# B$ J# ~
5 H; W( v4 ?# _0 @; `( w; A(a+bx) dC(x)/dx  = -(k+b)C(x) +s
/ Q: e( q) {- M( ^7 ^. O* M" l
+ W( s2 p; a  B/ [/ G* a. G
& P$ c( u: y: Wintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
5 }7 a. d* y- i: jwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

. v: E5 n5 }8 ?{(a+bx)/K} dY(x)/dx=Y(x)
# k. `0 {3 z8 b! x1 `( x% N! }$ ]
from here, we can get:
3 q( e' H. \# |  q" k+ r2 i2 s
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)

: N& o( r* V: T- E, ]9 Mthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
" g, ]6 Z  s3 l, @
-(k+b)C(x)+s = (a+bx)^(k/b+1)
$ N) e2 X4 {5 Y' u
finally:

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