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）
3 {' a: O+ c7 `
d [(a+bx)c ] /dx = -kc +s
& T) b7 j" I# h) q: M+ Awhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！

9 m) J" H/ L; l/ G9 w1 F+ l[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

' M, T! I- R  O. g/ Y(a+bx)c  = (-kc +s)x
5 D6 X9 ~+ c) w) lac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
3 J! P- V  @$ Q0 r
( I3 Y2 r0 V; j% D+ uFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

: x! C/ J' e% v1 Y2 l# i' \* lbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
3 I  F3 g* I; G- j6 Ri.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s

: B" R( T! f- Q1 G
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
! Y! T: c; S* e( h. ^2 S$ wwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

" L$ ^8 r/ O  r) C1 F# P; J# k{(a+bx)/K} dY(x)/dx=Y(x)
, R+ r% j& M5 b, L
0 U5 E* S  }- K/ N( g$ w# b/ Zfrom here, we can get:

. w+ K) ^3 H; z/ L8 l* G9 m4 U9 RdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
' u" I( i4 q0 [; Z
% v& a. `+ V7 U0 x* v, d5 A4 ^so that:   ln Y(x) =( K/b) ln(a+bx)
+ ]5 {1 g2 L  {7 Z: q) j5 D" t
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)