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
: @1 w1 J- U. ]; h% cwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

! y; V1 D, s2 I. E/ H多谢了！
7 h9 J  x5 E4 e  C: y
5 G8 Z* x! P& {% ~4 N* k) t$ f+ |' L[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

. |/ }% e, I! }6 [. r6 J(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
2 }4 C) d. w6 d3 @/ a(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
0 y3 n* x7 {+ Q
' S: M) J2 J. gFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
; G; H( V3 z1 C8 e. y( M0 Tso:
* {( k& s5 ~; _: q4 W& ?& j' J
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

/ C+ b! z- R/ V( a, ^% k" I, t(a+bx) dC(x)/dx  = -(k+b)C(x) +s

! Q1 D! r9 x" H; ~4 u/ l
! ~) f: ^2 y, T5 Mintroduce 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)

from here, we can get:
( Q. j3 v& K3 H# \( q
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
7 w9 \! E* D$ ]
) F' T$ I2 D6 E2 Iso that:   ln Y(x) =( K/b) ln(a+bx)
* v7 z4 j% P- U. w
  F$ e9 |* `$ b) L; ?this means:  Y(x) = (a+bx)^(K/b)
: s5 L) F/ q# b/ g; c, jby using early transform, we can have:

, n# \$ e; w2 H+ [- q1 O-(k+b)C(x)+s = (a+bx)^(k/b+1)
6 w% E' m9 ~9 c& e9 @
finally:

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