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）
, s, f8 n% J9 n) g/ v, L
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?
% |% K  Z( x8 O
多谢了！
1 I+ }2 T) |2 e. c, i# ^
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
0 i( d0 i! y5 U7 h, u/ I3 n* [4 e
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
  P  U0 o7 X( L(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
2 T7 r# z! C$ S$ ~' Q
$ g' G  j8 h9 @' ^! l+ iFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
. F8 J2 W: @& t. J& N2 @so:

: T  A% _% M  ~; u& ^/ abC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

8 q# }. ^, |9 L# n# _2 x1 w(a+bx) dC(x)/dx  = -(k+b)C(x) +s
" W2 _2 h" T, B$ ~) n) u* `

introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
4 w; V# E  a8 k; Qwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:

* p, G% W2 z1 X* U8 PdY(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)

& Z+ t! g6 B1 bthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
( ~* G7 L9 ~! [: q( K
-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

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