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
) d3 L& m& W5 owhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

+ a  D9 o; U" w- ~) D多谢了！

8 K/ ~6 C; C4 U, q$ w2 j[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

3 ~& ?# v8 h5 P2 M$ t& W7 n$ G(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

8 i! T3 m/ ~& \, B# YFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
! \4 ^5 b* E9 iso:
1 _5 ~8 Y+ |8 T9 `
) g% D: d# `# V1 C8 ~8 D) x# AbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
0 ]" U3 c7 ?( H9 z0 A" f

( o) [/ t( X1 {$ I7 h. k- }9 Kintroduce 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
therefore:

' V7 |: B) B+ E9 V; A{(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)

* [6 Q% M* [3 a2 X( _$ i' z  hso that:   ln Y(x) =( K/b) ln(a+bx)
( G2 |# E! E6 Y0 ~- |  B0 d; a
this means:  Y(x) = (a+bx)^(K/b)
" j8 D  v3 Q1 \$ \+ `# I9 u3 Rby using early transform, we can have:

6 b9 J+ b% \1 x7 G, w2 ]-(k+b)C(x)+s = (a+bx)^(k/b+1)
- K2 r9 u& Q- Y
6 `8 N6 I  d7 g4 V: ?( nfinally:
, l/ |1 \+ I( \( A0 f
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)