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
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

7 m5 x( U7 a, {& H1 E3 i1 f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
2 j% _' D: y, w$ q* h
4 o; Q0 h% z8 s* Q: K: e(a+bx)c  = (-kc +s)x
9 b0 Z6 z! C- j/ k) X# xac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
( ]; F4 _/ b6 z$ Y) w
$ Y7 U5 n2 i) {) c8 f5 lFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
/ M( j7 N7 r% |, n
6 M9 Z$ ^" i& k  Z: Z- W. ebC(x) + (a+bx) dC(x)/dx = -kC(x) +s
6 {' y  \  I" F) a" |; h' G& ~; ?i.e.
: [* {/ M7 |6 _
(a+bx) dC(x)/dx  = -(k+b)C(x) +s


+ t( D8 ?4 J3 W! |8 ~+ ^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
( v4 ~( P4 t: e- |& e( H; gtherefore:

1 d, J7 z5 N9 m- {{(a+bx)/K} dY(x)/dx=Y(x)
+ T, Z4 i" B2 K( H& @+ [. \
from here, we can get:

9 r" e( X/ Z. s4 o* b; ~dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

9 D$ L0 t  k' C& T, J! `& Qso that:   ln Y(x) =( K/b) ln(a+bx)

6 n* j% X1 J: P$ Gthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
" Y* a; m3 K4 ]
-(k+b)C(x)+s = (a+bx)^(k/b+1)
8 Y" Q8 F* K! C5 F8 {7 a* H$ F
  Z# l+ H# ^* O7 Bfinally:

  e" P4 Q$ \: l8 ^2 XC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)