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）

, W( ^# ?! p+ a" I2 K0 R& @& Td [(a+bx)c ] /dx = -kc +s
0 l$ c4 a: g2 f9 g3 Y( O3 t+ ^) N$ Kwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

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

* d" e+ |, J% d(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
& P. B9 q( J5 V) C3 y# A2 x& B; t(a+bx+kx)c=sx
; `! F9 `( S+ H( Gc=sx/(a+bx+kx)

十年移民路_

Solution:
# [6 H2 T( C8 S
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

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

. F9 S) y9 ~+ w; H! {8 U! E0 y; ]7 ](a+bx) dC(x)/dx  = -(k+b)C(x) +s
0 C/ T/ @1 O/ ?2 ^+ ?6 v
6 t( V2 ^$ z2 s! @- A0 ~# S, W
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
  }4 v+ K2 e0 N3 S% Y* Cwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

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

from here, we can get:
4 o/ ?1 x+ g: W% Q/ y, Y
' ^8 `2 l6 K. V1 VdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
" V2 a, X* x3 R
so that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
7 o( ^. J. P+ T2 S( ~6 j9 Jby using early transform, we can have:

: ?. c& v5 _3 s+ V-(k+b)C(x)+s = (a+bx)^(k/b+1)
- Y; a+ S0 d8 ?( ]7 \
finally:

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