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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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?
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8 q# `) {+ ^1 J: h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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* O% R0 V6 D6 a5 G(a+bx)c  = (-kc +s)x
, C9 a  ?# E' e- H0 O8 B* Dac+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
' ~. S" A* S2 l! a) lso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
9 ~# N) ~, w! C2 ~1 Ji.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s


7 t4 n; f( f4 pintroduce 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
8 U; Z5 m! T8 T- v( \therefore:
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{(a+bx)/K} dY(x)/dx=Y(x)

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)

" T* D% Q+ l6 T+ s$ ]9 w2 J; Mso that:   ln Y(x) =( K/b) ln(a+bx)

8 }! Z$ v6 v/ _% Hthis means:  Y(x) = (a+bx)^(K/b)
# R8 l& a$ {0 a% f. @6 wby using early transform, we can have:

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

finally:
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: o, g4 [$ E$ }& a. h, nC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)