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）
4 m' b$ l- `1 B1 ~
% u# h9 l9 H# J3 G2 Fd [(a+bx)c ] /dx = -kc +s
' }1 ?! W& V$ Ewhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

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
7 }2 t* f- w2 o6 M0 ~; O
. v  Q( O; D. l8 u- D(a+bx)c  = (-kc +s)x
$ N! W3 Y0 v  z, k/ C- Oac+bxc=-kxc+sx
4 q* R; [0 W. V(a+bx+kx)c=sx
4 M4 F* i/ v3 @8 `c=sx/(a+bx+kx)

十年移民路_

Solution:
) A# i. s" W3 ?& r5 U) f/ k
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

; l* g, L7 e% O' i* A# kbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
# O3 W* b1 Q1 Q6 o# w3 L, gi.e.
2 R* P* m0 v5 E3 X) {8 l" y' Q
7 q$ g" u( `0 C9 K(a+bx) dC(x)/dx  = -(k+b)C(x) +s
# K" z2 c- d+ V! N8 ]8 ?

introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
$ {* U9 V: y3 T3 a4 P( G2 k1 ^which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

) s8 y* C) d8 n, s* `! _* M' C{(a+bx)/K} dY(x)/dx=Y(x)
$ P1 T, d4 |  X+ F! y
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)
# r6 G( I( Q. }! G( L6 \2 T
so that:   ln Y(x) =( K/b) ln(a+bx)
4 U6 X. t. @: F6 R3 @( x
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
! G7 [6 P3 U: Q! D$ t3 X, v; \
-(k+b)C(x)+s = (a+bx)^(k/b+1)
: x. @$ V6 _$ _% N8 M" w* C% a3 F
finally:

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