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）
: U1 z2 X' d( f& n# l4 u) W" X! q- W
d [(a+bx)c ] /dx = -kc +s
# |9 K5 I! n; \* ^- O4 F  twhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
4 K3 i, `0 p% D+ B3 W3 p  A; f7 H
多谢了！

& O8 [: C- F  E% T$ A- S4 V' _[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

(a+bx)c  = (-kc +s)x
* p  t/ y' d( \" s- J: E: M: ?1 c/ Yac+bxc=-kxc+sx
% k5 y0 ]& t! e& V6 @" _(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
3 ?9 Q0 a( X; V& b' [
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
- V" b: O" |( t, a" |5 Q1 O# Zso:
/ z6 @+ o. T$ g# ]0 C
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
1 M7 G8 S2 Z. M7 v  L+ n. s
(a+bx) dC(x)/dx  = -(k+b)C(x) +s

9 c. z2 s( i/ L/ \* [( |
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
+ i5 E+ W0 J. ttherefore:
+ [! t+ q  u" M- U: @1 W
6 o/ D+ x) y( R6 C' ^{(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)
+ w3 A: H" Y6 X& i+ q8 o3 P1 R- c
* L4 E$ a* I6 L4 Y5 s3 @so that:   ln Y(x) =( K/b) ln(a+bx)
# B# S$ _4 w2 x4 j
this means:  Y(x) = (a+bx)^(K/b)
4 W; R9 L% }% i; A$ B3 M4 Mby using early transform, we can have:
1 d/ Y& U3 q! g& J' ?# ]
. `6 _! H. |- D! P) j9 r- b-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
  a/ Y; ?  e* A, X+ {; B5 @
, n! w4 I+ @! y( X. B/ w' wC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)