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）
; Q' t4 w$ H! R; B( s3 C$ l1 i
8 {+ z9 w  C3 a; L. e8 hd [(a+bx)c ] /dx = -kc +s
. s0 Q  W8 [/ `/ awhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！
& h, K: u2 {) I7 w
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
* t  p. d1 q( c% {
(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:

' F1 y, \+ g# \6 F% tFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

1 V5 I4 @5 U. I8 f* ~bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
- o* [- O6 }$ z) {: q
* O' l. H, a1 j7 x" a# ^$ {(a+bx) dC(x)/dx  = -(k+b)C(x) +s

. D7 b. x( R+ o& {# c
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
. c% ^( {4 l$ R# w9 \" {which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
: e6 |7 Y& T% ?! `/ I  ftherefore:
0 p, ^4 S: [7 `# g6 z  S# ]& P
{(a+bx)/K} dY(x)/dx=Y(x)

, Z2 a8 x' S8 _) a5 m$ ]6 w3 kfrom here, we can get:
' h  y8 K/ @4 C- w. m; v* M
0 I1 S; c. F+ U+ w; EdY(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)

4 {# Q2 N) j# I2 Sthis means:  Y(x) = (a+bx)^(K/b)
' @0 g/ M/ ^2 S, U& `7 k8 E# _- sby using early transform, we can have:

% }! i4 R4 p5 U- k% Q/ K-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
, h2 W) {7 s" I3 k9 w9 D7 I
9 Z) ]% J8 [1 H' ~/ n1 o- ^C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)