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）

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?
6 {' V1 j0 {/ e. ?0 f
3 e) h" t3 ]1 z) }多谢了！
  `2 O: r& o0 _6 H$ r9 d0 X
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
$ k1 B! L) E% g
/ o  a/ S" @8 C: Y(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
5 i0 ]( S6 q- W2 L3 b9 h(a+bx+kx)c=sx
" O3 @' Y2 c$ d5 Hc=sx/(a+bx+kx)

十年移民路_

Solution:
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# R  k  I" q+ h- NFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
5 K# y' `/ K; m, R, [
# z8 {% H0 @7 n: t. i+ [+ qbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s

* O5 T) G* J% a6 q% W  k1 Y, U9 F
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
  G, ^5 l# J! d9 L& a; w, Gwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
% c4 `( e8 o' X8 r- {therefore:

( U$ N  F* i* y{(a+bx)/K} dY(x)/dx=Y(x)

/ m7 A0 T! I' f( b3 ifrom here, we can get:

0 n0 [6 x) `6 e1 ndY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
5 f' _, S7 o5 h" i9 e8 l3 u# w
' v% H$ q; H5 [2 H% @/ i7 t! ^+ dso that:   ln Y(x) =( K/b) ln(a+bx)
0 c! F6 w$ w& G5 N$ U6 J) Q) V7 S
# A( \' A. l. J9 {' Y5 dthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
& a1 e4 K% ]4 n' s' B
-(k+b)C(x)+s = (a+bx)^(k/b+1)

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