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）
# p8 R* u; V3 g# d( n
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?
$ Z3 i* \# }* t# D2 o4 q7 Q( R! _! \
多谢了！
3 l( E3 G& C. n: z! E1 m3 x
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
6 ?$ R& X1 r  m" \: x
(a+bx)c  = (-kc +s)x
. \5 ?; q" i: A8 Nac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

8 d8 k1 I. Q6 O4 Y1 Y+ @From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
7 J3 f6 ^+ g  |* Q. V- Eso:

- x3 e6 u- ~( w1 d" I8 |; ^bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
0 H. c' N) M) d1 z! X. f# ei.e.
4 E+ x5 l6 K! ~" N) [% f2 o$ c1 W
(a+bx) dC(x)/dx  = -(k+b)C(x) +s


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:

7 N7 N# O5 Z7 l{(a+bx)/K} dY(x)/dx=Y(x)

$ M+ P  l' U$ V1 gfrom 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)
( J) p7 S; h0 o$ Z- Q& D/ n9 s
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

0 P" U; Y# d+ R# ^-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

" m# c0 B, @. H  W4 oC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)