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）

; T: l( |, ^4 ?: O# w3 Jd [(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?

! J/ W5 J" M! S- b4 |  M多谢了！
2 i  `+ W% V2 {0 u7 v* U* u( w/ g
* ^  \) c7 K, ?2 H6 e# p" U[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s

# `+ r) _& f4 i(a+bx)c  = (-kc +s)x
1 Z& N0 J) d' t5 {& ]+ t5 sac+bxc=-kxc+sx
(a+bx+kx)c=sx
2 _1 F/ q; u8 X2 gc=sx/(a+bx+kx)

十年移民路_

Solution:

* i6 G& m$ i" f/ @- m0 CFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
3 ~' A, k; E0 t  o8 Gso:
9 E+ z0 l( g8 ?$ r4 K: }+ i& G
8 Q; W1 u0 H4 l5 y! EbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
" c& c- f* u6 g2 Xi.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
9 v5 ?& F4 s- q1 u- u
: N" f9 z3 q# e4 _; k
. d$ S1 U- @' A. gintroduce 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:

( F+ x8 C  q5 ~; }" B{(a+bx)/K} dY(x)/dx=Y(x)

5 v: x" e" \# A# vfrom here, we can get:
3 h+ |! i9 M8 Y8 X) Z! j4 @9 l( {
+ U/ Q# S7 ^) I. EdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
2 X* P* O1 j7 Y
so that:   ln Y(x) =( K/b) ln(a+bx)
$ ]3 D- L6 @# [: c' z! c! L
8 L6 x9 U/ ~6 H6 F* t. y3 i/ Qthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
2 p) w8 `' Z# O3 }4 X( h1 Z
-(k+b)C(x)+s = (a+bx)^(k/b+1)
9 x! ~4 R" X. V: U- G6 G4 ?
7 n5 `$ s% h* R+ Yfinally:
6 E. H( X$ T4 D3 }6 O# B
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)