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）
0 K6 x# }% O0 S$ I3 Z2 G% _
# L8 G4 D5 @3 q$ G% y* |$ L8 od [(a+bx)c ] /dx = -kc +s
0 P, d8 J1 Y" o1 t4 T: Y7 B* uwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
: {9 S1 h2 X5 w3 {0 O8 I  S
  c$ m. {3 t" ^) I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

, N7 b! P) l( ?! v) e+ o! K0 d4 i(a+bx)c  = (-kc +s)x
; ]' f" u, h+ @9 ?1 rac+bxc=-kxc+sx
(a+bx+kx)c=sx
. q' b: n8 c. Q8 n/ Oc=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
8 ^/ V/ o6 I# i7 Yso:

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
! k& r5 c2 R! t. b( j

  q  b) R4 K. C" }( Y/ eintroduce 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:

" P9 o0 I: S4 d& M2 P; [{(a+bx)/K} dY(x)/dx=Y(x)
" K& l  k0 P4 `& N! @
  _' Q" e  c- R, t4 G) K# cfrom here, we can get:

dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

so that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
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) i1 t! O* b5 P6 B3 H-(k+b)C(x)+s = (a+bx)^(k/b+1)
. k4 T4 p- Y; ]) r: k
& s) c7 `+ ^; g" a4 |! Bfinally:
- h4 x. K( `  D! O( N
* ]6 K& Y  f; i4 _; FC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)