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）

# m) d* p1 Y! l4 ?1 G! D; kd [(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?
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, U! T% k) e- |" \5 ]多谢了！

+ i$ a+ g% C4 ^8 C0 x[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
8 S( q0 D3 R" \: G/ q" M0 U1 k, _
# b# j' D4 {% h(a+bx)c  = (-kc +s)x
" g2 f& o) P4 v/ e5 Hac+bxc=-kxc+sx
4 E; Z4 N( Z: }. I5 V) V8 a(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
/ e0 N8 D9 Y8 f5 q6 {4 h
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
) W$ F7 Q8 U; u& ui.e.
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9 _3 i3 s- B$ ]5 ^(a+bx) dC(x)/dx  = -(k+b)C(x) +s

/ r% J4 H6 g' X$ ~" v  f
5 d) n5 ~. z$ S# T$ \6 cintroduce 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:

{(a+bx)/K} dY(x)/dx=Y(x)
- @( m- i4 \0 A% P' x
from here, we can get:
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; q2 c% I9 F) T3 o; c. [dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

# y9 \; r4 S/ ?( Bso that:   ln Y(x) =( K/b) ln(a+bx)
2 b  [# k( K, X. X" T. J
this means:  Y(x) = (a+bx)^(K/b)
! N+ D7 w* ]6 I3 T' Fby using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)
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5 N) T  b3 X! ~! [finally:
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$ p5 L  ^( e6 o" U4 P0 X. B" _C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)