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）
: Z6 \0 ?0 ?) H3 l- z( a
8 Z/ K+ p% v0 yd [(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?
% ^$ ?3 G5 U' W
多谢了！
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[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
* k( b- w: K. K' G! j; s( B5 Q
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
# M1 L1 _, g/ t3 p) }7 ^c=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
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bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
- e" o1 I: C: P  [) K  k2 r6 ki.e.
# V$ D6 u0 ]. A  i$ f0 Z
/ J7 G& t0 L+ v(a+bx) dC(x)/dx  = -(k+b)C(x) +s

- S8 Z2 D4 X) U; A# I
introduce 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:

{(a+bx)/K} dY(x)/dx=Y(x)
# p- H7 n$ m6 y2 h# w
from here, we can get:

4 c5 Q7 p4 J( {/ C+ k* CdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
4 d8 G4 w9 K) F  U0 e
9 c$ F* Z$ H" `; J9 K' E* Eso that:   ln Y(x) =( K/b) ln(a+bx)
3 S  c$ ~4 l: n: f2 J5 F
& @7 a8 v  U0 Q* ~& Uthis means:  Y(x) = (a+bx)^(K/b)
2 V9 |; ]' e" Bby using early transform, we can have:
0 ~/ a9 _( B% b! f  X
( j' I9 F9 s7 @1 o-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
: v# u4 C( Y9 Z7 |* U3 w
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)