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）

' b5 x( F/ Q! b2 Y* [- H( zd [(a+bx)c ] /dx = -kc +s
( l3 r% H/ E5 W5 ~9 wwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！
: X, |3 d( z4 v8 c1 ^& u2 M2 c
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

(a+bx)c  = (-kc +s)x
' G$ w+ U+ U9 `  o* kac+bxc=-kxc+sx
(a+bx+kx)c=sx
/ z$ X& D  Y/ T$ {' tc=sx/(a+bx+kx)

十年移民路_

Solution:
; N3 o) x, |; f  g0 c/ s
/ j' N, ]# n1 _6 z9 [From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
# X* p) W/ v4 l7 X- G' I
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
6 T/ u9 @/ {, y3 x( e# H- _
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
: T. Y! R; Z: @8 ~* j# g; S! c8 w$ E# [
" f5 l( y2 T% ~8 z8 \9 n
& e/ x/ [: K$ Z/ {% k& @introduce 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:
( H& R/ d& g# Q' L' w9 T8 M
{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:
  P8 r. B$ f/ l+ f  D5 O4 v
  G' C6 I# X6 _' _/ P. H7 A4 y' IdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

1 K* U3 ^0 r$ d2 p8 yso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
! y6 T+ o2 D4 y1 }3 fby using early transform, we can have:

& n- C* e0 H$ v% Y, d- O  @! D-(k+b)C(x)+s = (a+bx)^(k/b+1)
; {# u  M! I, c& p- w
0 h' M( Q3 k+ \5 X0 M! `! Nfinally:

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