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）
0 f+ L, {* h: [4 e' Q. H( ~1 e
9 }  U- p: Z& a) _9 N3 sd [(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?

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

/ D9 N) L) h# x8 ]6 k* p(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
/ l" Q4 J% Z" o( K5 v+ g(a+bx+kx)c=sx
( |9 r* [; w! p4 jc=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
/ L3 r" b" @( c! o, p3 y; pso:
( g. c  L( N; P; l" d( }
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
6 Y9 b! e: W; ]: \1 ]( M9 i
. r7 L9 ?; a! K  ]; m8 }(a+bx) dC(x)/dx  = -(k+b)C(x) +s
2 H& u2 ?3 T3 G; n& u
+ U0 B  c& z% |# ]+ r2 ?9 I
* U$ t3 k: _# v; O5 ?. y0 L" _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:
# [! Y3 S: V- C2 |* n* }+ }$ P1 N- u& h
) {7 B: V$ U1 o! i{(a+bx)/K} dY(x)/dx=Y(x)
% o. y4 r# @$ P
from here, we can get:
  P; p5 _  ~; n) A5 m1 k
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)

this means:  Y(x) = (a+bx)^(K/b)
2 @& B! k- g2 j! @% d! R) ]* Mby using early transform, we can have:

+ B! l) G0 f; }& S-(k+b)C(x)+s = (a+bx)^(k/b+1)

& m9 W& K; f4 }. Y- A2 ^0 Qfinally:

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