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?
9 [7 j2 T, M9 s* u
多谢了！
7 r/ ^5 U" M$ x' M: Y
) J4 l& W! F- R) X& j[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
+ o4 ]+ {/ h  K. L6 K
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
2 S  @1 q& c# w. f0 rc=sx/(a+bx+kx)

十年移民路_

Solution:
- d+ S0 G/ R2 s' @. N
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
7 }: v1 f6 t( J0 T2 s% Cso:

/ F  c) y$ I" ^bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
4 F% B5 B! V% E3 ]! O
4 D7 ]$ k. ^. U( [& t- a(a+bx) dC(x)/dx  = -(k+b)C(x) +s

, [* P( `& V! S: J
+ ^) g0 p2 w( [introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
0 R: }, P+ n3 k' A8 Twhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

" x$ |2 O7 j- L) W5 B& Q{(a+bx)/K} dY(x)/dx=Y(x)
# Q7 k6 @/ ?: l
/ j) F- _- ?0 z1 \3 y" \, }& Ifrom here, we can get:

& q3 Z! B$ }9 Y9 X0 g2 qdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
- P1 @6 N4 E! t) Z$ y  {
! o) R/ w' _1 I# ^. R; H) Uso that:   ln Y(x) =( K/b) ln(a+bx)
6 b3 ^* s1 {* P
" t# t6 V% N. ?1 v( C6 {7 \this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
' m, v4 [. C4 E
-(k+b)C(x)+s = (a+bx)^(k/b+1)

( `3 P* v5 s$ J& y+ wfinally:

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