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）
# Z. l) @6 K2 ~3 g# O
" G% i( e" u0 d  rd [(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?
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# ^' }. M+ X7 ~" U& E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/ d9 J* I2 u2 A7 i: V" o! J(a+bx)c  = (-kc +s)x
8 r9 s2 v9 @: I2 m+ |/ J& S( _ac+bxc=-kxc+sx
# D' O. y$ t1 B. \8 V$ \(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

3 t3 o# L) d# ]0 x7 b6 V# o2 i: GFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
/ u# r8 _/ w. e: a/ G, G# Z
6 e. u( [& }- D) i% ^: v
5 h. R2 [2 ~2 k" X( o, |introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
4 M0 ]) V4 c3 J; ]  `which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

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

- p# i5 }  e# {8 E) Afrom here, we can get:

, A* B- d  e* b, u0 _dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

# ]2 d" D8 N1 i- v7 X" ?) D: a. kso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
3 s2 {1 [- a( n! `by using early transform, we can have:
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-(k+b)C(x)+s = (a+bx)^(k/b+1)
# ^3 T# P0 u. T+ e5 S1 x
# g/ l) V  K- W. }; w& F4 Afinally:
1 {! a* w/ ^. ]! u: k+ l) p
% q# l. S; A9 {C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)