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?

多谢了！
& s# R$ |+ n, e9 M
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
  g' H- U6 u! c4 u" M+ D1 J! P
7 s" ^8 g, Q6 W/ W9 T! M0 s1 L& c: h(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:
2 [, c* z' \) K4 p$ i+ A) o- n
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
$ c; o& N. T& Z1 G9 `" lso:
5 X9 A  e9 O: [& U" C
, @- `7 ~* `( WbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
# q2 s' w9 x; b7 f% a# y- F9 wi.e.
& w' J' k# j! r& ~
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
  [" v9 l% V8 |" X' b/ r  [& {
& `% H* O$ h7 c& \, S+ S9 ]
# }+ a+ o% M! g1 v" q0 {2 E- Iintroduce 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:
; S; v& `* [! F
{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:

# T) Q/ A( N0 f; ?( L& g! W' s4 R$ z: VdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
- R7 v  _7 f8 S) M" ^
' I- D3 P# g$ i' c+ ^0 Yso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

; ^# `) `: v* a# n6 B/ I) d5 q9 D-(k+b)C(x)+s = (a+bx)^(k/b+1)
: R9 H! n0 Q# l; b
finally:
& q5 f5 k. a. \8 {/ f7 e
; Q; p: Q7 a' ?! E5 RC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)