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）
0 M7 V% ?; s) P0 f
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?

多谢了！

[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

4 G" A7 [& i: S, q; M* k& b1 i5 K$ ^1 _(a+bx)c  = (-kc +s)x
& O! _2 |  d: p' X. I& Yac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
; a* {+ w. M; C1 kso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
" Z: ]- O) u$ [' y
# g9 e1 Y2 c, A2 F0 i(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
$ B6 \+ S6 s* v; I7 utherefore:
& ?0 |( e) h# \# {1 @6 P" r
{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:
  i4 n! ]8 r3 I2 s$ W* |9 |
( @3 h5 \: H  B3 T3 |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)
. w3 h& b# M4 F# y  T9 o: @
/ h7 R* f. B3 P/ Xthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)

; s# d; a4 V1 Sfinally:

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