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
$ ?9 g) Y  P* u4 w. t" m$ Xwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
& W5 K  x' i3 n% W) U0 M
0 r; ~+ d; S1 B* y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

: I. b% t# W& A8 v  K# ^( e3 Q: ~2 r(a+bx)c  = (-kc +s)x
9 R1 Q1 Z* E, ^1 `ac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

. d+ z8 G+ W" `3 X1 X$ T2 Q$ @; HFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
3 H0 x1 ^* ?3 qso:
' R% M- b/ M) I1 Z/ a) D' ?& r
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
! |' M% }3 W, E& Xi.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s


. x( Z- P4 ?( t5 T6 }' V9 Iintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
$ L6 @2 w+ r3 i! k. pwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

2 M/ I* [: k3 w. n2 c5 }" D{(a+bx)/K} dY(x)/dx=Y(x)

; q8 N4 i- G" D# M. Qfrom 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)

so that:   ln Y(x) =( K/b) ln(a+bx)
, m# \/ y! q, ^' _
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)

; F, ?- d3 ~) u3 x6 m4 Gfinally:
6 c* b3 S  o/ D" ?# a5 O
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)