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
/ @, A) L5 f0 k9 v" n) `1 f4 V3 @where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
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: J! q$ p7 [8 G: i) m多谢了！
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7 t7 d0 K/ L  P2 s8 w; `[ 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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" {2 m$ Q- _" |0 U0 Q9 [* R(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:
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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
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( ]! B8 {( i4 w) |- J(a+bx) dC(x)/dx  = -(k+b)C(x) +s

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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:

{(a+bx)/K} dY(x)/dx=Y(x)
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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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. |  B. y5 }! M! uso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

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

. W6 b. i6 e  H9 J% _finally:
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0 [" R" E6 y, W  |& |C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)