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）
2 C$ Z+ E6 S) m6 C
: A* J) y$ L# y: Bd [(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?
( m/ g; X1 }8 H4 r6 o+ [- v
3 @- r) W3 C8 J: ]7 r: J3 I多谢了！
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5 z8 t% N  v8 l[ 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
5 T' Q' {5 o' y$ n5 @9 e% T- ?* c(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
# M, `: n3 ]; d" v
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

% J" U; F) m8 `/ E* l& U. LbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
$ }* Y5 d$ i3 p3 i+ x0 c3 @! yi.e.
8 v' d6 N( p0 {2 R
(a+bx) dC(x)/dx  = -(k+b)C(x) +s

" x4 U/ [2 V5 w2 K
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
. t# @$ e( r' A6 awhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
/ F  V) k6 ]# n0 \4 Ptherefore:

0 ]1 @" W- B) n+ U9 J{(a+bx)/K} dY(x)/dx=Y(x)

9 |# y/ \! v& j2 Ffrom 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)

so that:   ln Y(x) =( K/b) ln(a+bx)

& t+ V4 o; K6 T. |+ U) qthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
' ~& H, E: ?2 J5 a4 ]5 w* i8 w* F  |9 z
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)