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?

6 E4 o: C5 H& g多谢了！

( N! x' m7 N- u" W[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
/ X+ N+ o4 V( y7 r
(a+bx)c  = (-kc +s)x
# \% Z$ X4 D3 B4 W% L- |, R4 qac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
* f1 i9 N* R& w! v
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

1 g- e# f6 q7 [! g) L% ibC(x) + (a+bx) dC(x)/dx = -kC(x) +s
3 a8 r4 _# ?  ]  o/ [i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
& w  g" B+ s3 J

# p; j' H9 T4 I" B2 @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:

4 v/ S7 x3 d. K/ R. |9 B$ _{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:

, m1 D% \  H) h3 y: g3 w6 D+ ^, ~dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
- m' L$ q. Y0 X, x# c/ A! J, U! T+ [
, j' v/ l4 h+ `! k: aso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
+ _0 l6 N, E9 pby using early transform, we can have:
( e& C) o/ M) ~7 f0 g
- w$ D& o2 B( c  a-(k+b)C(x)+s = (a+bx)^(k/b+1)

# a6 L0 w1 ], }$ P. cfinally:

" f& e' {) d% C# ~# PC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)