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）

' S1 V# G0 {: n4 i6 i. Rd [(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！
; l/ _6 v: V7 }
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
" g- N1 }3 m9 ]  A
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
3 k7 F. G, B3 l- g$ k(a+bx+kx)c=sx
% s1 v: g; ?! p, H% Uc=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
: @1 e" ?" S* G+ B/ z
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
/ j% M  F" Y+ u8 ri.e.
# A. Y: g) c" c9 h3 H! h5 n
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
" d7 A6 v/ |& x% \: Z/ S
/ i# n9 n% }# J, V- p
+ O3 ], P& A% Z0 I+ y$ o, sintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
% a* w+ {- V: i1 m; u& G/ Pwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
: Y0 l! ]7 @& S& L9 utherefore:

{(a+bx)/K} dY(x)/dx=Y(x)

+ w! w# N) X, p8 Efrom here, we can get:

$ P* K. J+ @. ]. U  P# sdY(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)
4 X. l$ L6 i; ^0 O9 Y7 T7 ?& [
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
( H) N. l9 ]3 Z0 S0 F
" L7 b& J# n# d/ h& j- |7 \-(k+b)C(x)+s = (a+bx)^(k/b+1)
' h+ z2 B. ~. j- w* C6 c3 _
6 i; q1 A: X5 d! Z6 f$ tfinally:
9 i+ A* I7 x( J, n6 T, K* Q0 y: U
* x) @1 w" |8 n- \5 Y, R- X9 e1 {7 gC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)