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）

; h- \+ u; `( m. v  E2 w1 B4 l! ^, R( ]& Nd [(a+bx)c ] /dx = -kc +s
( u" n# _! V4 C$ Owhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
% r3 t' d7 ~% D7 B- J, y
多谢了！
0 H/ W5 k6 a4 c, z2 w- q, n
  {7 [+ T  X5 n  r[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

/ z/ W7 h: ?. G(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
1 k. r/ w# g1 e" Z& T$ i(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
& a; k  @% q! S9 k; B# rso:
; e2 _1 J! g7 K7 L  N5 [
+ m" g/ W( Z  b8 _/ f4 qbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
' `9 C0 r$ ~! P; Q( n+ E& ?
9 u& v( {. Z1 t0 z(a+bx) dC(x)/dx  = -(k+b)C(x) +s

8 a2 \* z7 h$ M7 z9 \0 d4 o* h: h; n
+ _- X8 T9 [" Pintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
, C% [2 b, q; Awhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
4 {- ^1 D( W5 @, n  m4 |7 K
{(a+bx)/K} dY(x)/dx=Y(x)

2 \  W! |+ f' j# e' \, Bfrom 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)

" R, c7 @8 j4 |6 k4 Ethis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)
, Q* U) M" s+ f! Z! ^& \: E
& ]) Y: z: q5 P7 `$ l, Sfinally:

/ e4 y: q& B) x' i9 C- e3 a: IC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)