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?

5 l& u% z( O! e! m5 c多谢了！

[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
( S" z8 Y9 C+ {; \% h' B
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
" t2 Z! ^5 h  L$ Z+ w+ zc=sx/(a+bx+kx)

十年移民路_

Solution:
0 T1 H7 n# \( ]
( f# ?) I( f$ j$ w' IFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
+ ?% S: T. _/ F, w6 H! B7 G# [
& [: G4 t- M4 sbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
1 J/ n. D7 S1 wi.e.

: ~6 T1 O+ [0 k& M, D9 Z! C. G$ Q- S(a+bx) dC(x)/dx  = -(k+b)C(x) +s


introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
; a4 Q+ m, q/ Twhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

( z. R$ G0 _; }% ~' G1 K- l{(a+bx)/K} dY(x)/dx=Y(x)

5 }/ L. a7 w6 a6 l$ p/ G" x4 F' A) Ifrom here, we can get:

) a, J9 }0 L  N2 s, S8 Y# A3 ~. mdY(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)
3 s; J: u1 Q# S% R8 W, a% @% C' ~
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

/ Z4 I+ E% t" c" r& k0 y, ~-(k+b)C(x)+s = (a+bx)^(k/b+1)
! z& i8 t0 O- R- ^
9 C+ N" b! |! E; s7 C1 \( U, B2 ifinally:
' k5 h  ?4 b. s( x  t/ v' z* _
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)