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 n3 K5 Q% I; z8 Y8 p* ]5 V! a% 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?

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

5 v7 t# C  S5 |# _: |' S; Y' C$ x(a+bx)c  = (-kc +s)x
# d; p% c! V( z5 m4 x3 j" ~+ Aac+bxc=-kxc+sx
' ^7 ~! X, F) r5 ]$ q5 G(a+bx+kx)c=sx
, y4 q8 C' K+ }; _$ N4 Q8 Qc=sx/(a+bx+kx)

十年移民路_

Solution:
0 x* M& J' X) u. X% R
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
9 O* c! ]5 z4 x& _: Pso:
& G  B+ o- q; p2 }: ]0 I, G
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
& ~( ]- w1 P  o$ H) V0 d- {: O3 Ii.e.
2 p7 V4 m0 j+ S$ m1 Q
) j7 I: S& r8 D- S(a+bx) dC(x)/dx  = -(k+b)C(x) +s
; g' q) N$ N% {
. M* @1 b' d' v0 I) V& r' o: F2 Z
& P4 Q! z8 G, g: P/ ~introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
! S. _# ]: n9 }3 m& H: |which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

+ R& \& \6 [1 N+ n* m: ~{(a+bx)/K} dY(x)/dx=Y(x)
. J# @8 f- H, U
from here, we can get:

dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

5 H% h1 q  x5 m6 G% U5 |so that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

% ?& K0 x0 o, }-(k+b)C(x)+s = (a+bx)^(k/b+1)
( Z' D5 l% E2 o4 S( ?. _4 P: G
7 G/ c% ^- K, j  T0 Pfinally:
; p9 ]9 t1 `- r  l  w& I
2 V3 F- D3 g" C3 g4 u$ m% jC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)