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）
* J. I& g6 w% v+ `
' h2 j; P/ }; y. o, R7 x( v5 Dd [(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?
" s0 J0 \# b2 u: X
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

(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
. `) z4 F+ ^: K1 G4 f# ]& Q(a+bx+kx)c=sx
8 h- F" {, H  A1 g1 Fc=sx/(a+bx+kx)

十年移民路_

Solution:

8 Y- @) \1 _+ `7 @: X0 \From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

+ }& X+ K! v7 i8 z' N* S. g$ `# x6 a9 abC(x) + (a+bx) dC(x)/dx = -kC(x) +s
5 }1 R! M7 E% gi.e.

: j1 K, o# P. Q1 L(a+bx) dC(x)/dx  = -(k+b)C(x) +s
+ k8 c$ k' B& i

introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
+ Z6 z7 X" H2 Twhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
5 z( L# W6 _- d5 K0 q- ktherefore:

{(a+bx)/K} dY(x)/dx=Y(x)
# f3 O/ f$ q* L9 L2 x& h4 E
7 i5 c7 _7 ^9 X  c0 Y$ nfrom here, we can get:
9 x( c( X/ }# c
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
6 p4 C$ i( H3 k/ ^. L( w! H
: `- b: G6 P9 mso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
/ ?0 ~7 M2 g. u9 iby using early transform, we can have:
& m# b+ }, U( A, i
9 k. r3 w1 [- ]-(k+b)C(x)+s = (a+bx)^(k/b+1)

( f) r& R' c" b$ c$ [3 jfinally:
% X; R5 S# r; t0 `, ?
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)