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）

8 F/ n+ U6 S/ Q( Md [(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?
$ v1 f* `) b8 L/ C
多谢了！
9 w/ n4 {0 g% g2 G+ e  r) `: T
[ 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
, u7 n4 \& }$ I(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

) w9 j. I  r, nFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
; _+ h/ F, k7 c& E' F, o; ti.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s


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:

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

from here, we can get:
8 ]: o* q3 X% V
: B- r5 l2 C3 b0 a# wdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

# D- G# ^( D7 R% G) _4 Dso that:   ln Y(x) =( K/b) ln(a+bx)
3 h$ a+ H, |  G. W1 r
8 Q* x1 ~5 h5 U  ^this means:  Y(x) = (a+bx)^(K/b)
. v6 _! d6 ?- ?% [; a6 Iby using early transform, we can have:

( t0 A0 @+ |7 B3 Q% Y& K( j-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

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