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?

' b/ a3 j6 h; B8 v多谢了！

# o8 A0 a- U# S) c[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
3 A5 O. K; G8 u' R) q/ \
0 e* X- e% ?1 C5 K(a+bx)c  = (-kc +s)x
% v, y  D6 r/ O" P. Dac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
# d! B0 n; E. O! q. c  U
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
6 h+ |! k, ?1 Z! X1 _  M
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
0 {2 [( ?3 h. v- u8 Ti.e.
- K5 n6 U. ], p/ q; S* {% z
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
* u2 Z$ r( t4 Q4 e
5 ^: C, k8 L+ q. J
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:

: E; o, C6 M# }" K! L1 w2 @{(a+bx)/K} dY(x)/dx=Y(x)
! h5 j0 s, j5 [) d" N
from here, we can get:

5 U. U0 i: r, P" J- c) gdY(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)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
( `- X/ C; g8 J( v
7 F2 @# l/ U1 ?9 w( P-(k+b)C(x)+s = (a+bx)^(k/b+1)
$ C: Y) y; e4 p* t, K" ~6 j
finally:

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