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）

$ B" {6 T& n" t% ]/ Id [(a+bx)c ] /dx = -kc +s
" N7 I4 w# v! S2 B+ S1 l) i; D, E6 v" A; Dwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！
7 U! z. C# y, z  K0 d* b
[ 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
7 W- N' z# n7 u: h, G(a+bx+kx)c=sx
# l" l* A4 a! h( w& vc=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
+ R: X2 L8 a$ G1 k. E6 c- ?  rso:

2 O9 D7 g% ~# `bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
$ Z3 T, v( O! J3 l" F/ e

introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
4 g2 n6 I' v/ d7 M7 W) h0 F: Z! Nwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
+ V4 U' \0 g0 [5 itherefore:
8 D9 Z4 b( U( C2 W/ ~& N
{(a+bx)/K} dY(x)/dx=Y(x)

3 d% a' R/ [6 ]: ~1 }7 D! e( ffrom here, we can get:

" c2 a1 K, R2 R5 }dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
3 b: I$ h6 b' I  q9 o  b: w9 O+ w
$ v  S. h) ?2 }2 W5 Yso that:   ln Y(x) =( K/b) ln(a+bx)

5 \- [( n  p# [4 ]this means:  Y(x) = (a+bx)^(K/b)
/ |6 m* q# @2 _. k6 _by using early transform, we can have:
1 j2 {. i& a8 `3 _5 G+ D+ R/ w3 v
-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
; p6 z( S6 h! H. ^' I, y
: Y! \, ~5 z, M  j3 J5 P7 MC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)