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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! k# F5 D, Q# w3 k& c: ~" m' td [(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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0 a5 |& X3 A& M, ^3 I7 u  C. c4 b. 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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(a+bx)c  = (-kc +s)x
( v# s7 s6 _# x; Oac+bxc=-kxc+sx
(a+bx+kx)c=sx
9 }' L0 H: D! d- c  _4 d/ Bc=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
i.e.
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(a+bx) dC(x)/dx  = -(k+b)C(x) +s
3 {: }  {+ \1 U
8 P3 c, T1 f7 k& `4 h/ C* R& ?
) v; E' j5 |6 p) s" |introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
7 E. d& }% ^; m) p, E; U" fwhich 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)

/ Z6 ]/ Q/ W3 V8 d' r. sso that:   ln Y(x) =( K/b) ln(a+bx)
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this means:  Y(x) = (a+bx)^(K/b)
! @& I: Z7 Z7 c* C1 Z8 |9 y/ o- g2 Xby using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)
2 u/ k$ F; z8 R0 B3 k
finally:
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C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)