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?

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
* t! f, h, `8 K$ T) {* q* A
(a+bx)c  = (-kc +s)x
- G( J; W' {" P' U+ J* ]. O; xac+bxc=-kxc+sx
/ `# Q* @1 M' q; v* Y- J(a+bx+kx)c=sx
% [. G" z* i& `c=sx/(a+bx+kx)

十年移民路_

Solution:
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1 J% V! f9 g- h; {5 }+ DFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
- F5 x9 V# Z" f( Y+ \5 ?so:
7 Z, b. \: ?9 d- h2 {% v
* T: a+ n9 d/ |! \  pbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
$ N2 V! v1 ]& L3 G& c& U
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
* h" S2 p6 A5 R5 {

& C, s! x3 u8 r: E. K$ K0 ~introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
" u% H0 {+ G( s1 B5 Twhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

! w4 \8 r0 w) i4 ]" Q9 l{(a+bx)/K} dY(x)/dx=Y(x)
6 g+ d  Y. q3 J7 @/ \7 W9 j9 N
$ I" E/ G( o& J5 t  Bfrom here, we can get:

7 }6 U2 y5 r  S9 x/ ~dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
# ^6 l( w7 z" A  S) I
so that:   ln Y(x) =( K/b) ln(a+bx)
. Q0 W3 g4 I; h% D' i
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)

3 [6 z! E% D% i) pfinally:
0 \+ n" A; x( |" T8 Q
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)