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
# q0 i. y1 W( E8 `1 [% H  ewhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
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多谢了！
0 `* w6 A8 R- z" ]# t
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
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(a+bx)c  = (-kc +s)x
ac+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
so:
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bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
- ~; h6 j' I2 v1 u# v3 Bi.e.
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% {7 E" T, h7 \% _(a+bx) dC(x)/dx  = -(k+b)C(x) +s


introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
( O7 w+ h6 e& h$ w: N) iwhich 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:

1 s3 r# q4 n0 Y- WdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
+ d3 m' {' |& ^! ~( }4 t
so that:   ln Y(x) =( K/b) ln(a+bx)

3 }' _* c- K9 T& Dthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

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

( |9 M- {' U' ]; Sfinally:
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/ f5 K9 P3 |/ W, c9 s: GC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)