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）
/ o3 K2 o1 @9 ^" d2 ]) J. l
d [(a+bx)c ] /dx = -kc +s
; }; n3 I/ r$ s% s0 K+ \9 Bwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
9 C: E3 f% ]. j1 B
- E* h2 i0 A5 H; H5 y多谢了！
9 M! }5 j/ _( r) e, B: V9 ?7 P  Z
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s

) N( m0 h! T8 M% ^(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
+ j, ^5 k% L! M/ f$ Q2 ]' [/ k' {c=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

7 D* o' D+ _6 Q+ {* C(a+bx) dC(x)/dx  = -(k+b)C(x) +s

. t$ t4 V; S" }# \: ], k6 L6 h. F
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
; S" j$ C$ o  \8 d0 owhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
: G% J2 D* s6 @, W. D3 Y8 @+ Ktherefore:

- @2 F# i9 U4 v, I. a$ z{(a+bx)/K} dY(x)/dx=Y(x)
, M& A* G, t8 I/ E. y- X5 _
  `. M6 g0 e8 i3 \. r3 n9 ^from here, we can get:
0 \3 s0 k, @' Q
/ l9 C3 }0 U, F* P) b; h' m; {- wdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
( I0 ^. c! C8 E1 N
so that:   ln Y(x) =( K/b) ln(a+bx)

+ g4 m3 B1 U* w$ j( bthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)
5 j. j: G: q% _5 z
) _0 {3 i0 E* Jfinally:
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$ n* |) m" U! M7 Y2 T! bC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)