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?
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. G- V* J; t2 r( ^0 n* v) k* l$ S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
1 I! ^' ?4 O9 I/ g: i, s. S# l) D
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
& n; B1 f, U9 n. Y  U(a+bx+kx)c=sx
$ B' Y2 W9 j! g4 k! J. ec=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
* C4 S* Q5 r9 Z+ a: J+ o+ bi.e.
& N1 t+ q$ `* X2 w
(a+bx) dC(x)/dx  = -(k+b)C(x) +s

4 l! I5 L& Z: }6 @
  V! Z7 `1 w+ |4 ^9 {introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
, b0 G& |/ r& Lwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
; _1 \' q/ b; W6 }  p, L( e3 atherefore:
: Z7 g3 p7 X  ?& k
2 _- l, _, K, }3 |8 P{(a+bx)/K} dY(x)/dx=Y(x)

2 f/ N3 b" V* q, }. jfrom here, we can get:
2 J9 q% p5 q) j/ T: [- p
6 z. ^7 g7 J5 y+ RdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
  o: U, ]  g6 `8 s5 y6 ~0 t+ [
so that:   ln Y(x) =( K/b) ln(a+bx)
, Z3 \  e% Y! z- |% e- W+ s
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
4 D" H% j6 s/ y% s8 F  L/ l/ }" d
-(k+b)C(x)+s = (a+bx)^(k/b+1)
: M2 I4 l; r4 q+ y
5 O$ d/ T$ \8 V7 {& H& l7 lfinally:
& m! c8 _& x' B3 C
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)