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）
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, L- R3 w0 f3 d9 w) w  h; D- I! Xd [(a+bx)c ] /dx = -kc +s
  q' W4 o, {" a% J& A$ c( mwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！
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[ 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
5 ?, }: j( ]1 N" lc=sx/(a+bx+kx)

十年移民路_

Solution:
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3 x& g. ~& a# i* t, Q4 |From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
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: t& u7 J6 L! d$ TbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

% r( D5 E( b6 K5 a" [, @(a+bx) dC(x)/dx  = -(k+b)C(x) +s

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! V. @4 |" B9 A1 o& i( a1 Hintroduce 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
0 E. M4 i7 W: ktherefore:
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{(a+bx)/K} dY(x)/dx=Y(x)
3 O' T5 U2 I6 w9 l3 [% ?/ {3 d: ?
from here, we can get:
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dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
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so that:   ln Y(x) =( K/b) ln(a+bx)
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4 W2 h7 d4 s4 D, [! y+ }this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

2 X" e; i% i3 N1 K-(k+b)C(x)+s = (a+bx)^(k/b+1)

4 \9 P- j3 K: h/ \finally:
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C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)