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
* H+ [5 [+ |! R0 s& }/ E+ Jwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！

+ s6 w: T4 |6 S! c2 g0 u4 ?  [[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
7 g( ~) L3 w5 _  H) V4 \' ^& F  _  n
(a+bx)c  = (-kc +s)x
4 G& `' }: j5 |( Gac+bxc=-kxc+sx
6 F; G7 h- ?$ v! R( J6 {' T(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
; O& K$ U- T2 q4 z) ]so:
8 y- u5 n! L: [) c" R
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
, o6 }- H" s, ~9 Bi.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
2 X* W5 C: y% s# _+ t" C! Q; y
  ]' P+ C, E( I/ n  w$ |
introduce 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
( f2 W9 w) B9 @; w9 f' y* S* stherefore:
0 U; I; C2 d4 G- n1 m
" j: q, T. l  B8 {4 B{(a+bx)/K} dY(x)/dx=Y(x)

) J! C/ }+ `1 U5 afrom 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)
2 f. N3 A- Y8 t, }
3 f' `! c; y1 X/ n/ c! {/ dso that:   ln Y(x) =( K/b) ln(a+bx)
! z# ~6 V1 H) j5 t5 o( }
this means:  Y(x) = (a+bx)^(K/b)
# T1 _; A2 U3 x2 b: n) Bby using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)
3 [1 b! w3 U1 U4 A6 Y4 d: f' V1 S
finally:
, h7 `2 B$ C. K% {" I2 c$ _
& T$ o, P8 V: J: x$ ~  PC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)