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
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
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2 c/ ?# a/ o! m" X" W6 i多谢了！
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8 @* r7 L; J( b" ]: y! u1 D[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
( E2 K8 t) i1 L2 d* S( l" G
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
7 {2 ~" T/ I4 F+ a4 }  @) ~" p1 V& R(a+bx+kx)c=sx
4 s$ N& q  k# i. X9 O% \+ ^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
$ E, G# y! J2 ]& u% y# |so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
; K3 n$ E5 S2 x
8 e, h, }: s7 X$ r(a+bx) dC(x)/dx  = -(k+b)C(x) +s


1 {1 @8 `9 |" q6 uintroduce 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:

. m+ F! E  o7 z; [7 Y3 E- f: E{(a+bx)/K} dY(x)/dx=Y(x)
" k) B4 O$ ~8 C+ ?; j/ j, Z
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)
4 Y  @) }0 R3 ^3 j
so that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
8 ]3 Z' E( N6 o, X) j4 ]. g3 Wby using early transform, we can have:
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  L5 P5 }  M( h, Q0 d7 P-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)