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）
" }; E( `! a1 h$ c
) G* C" h' f) w4 xd [(a+bx)c ] /dx = -kc +s
# Q" V, V5 L2 t- f! u% H3 L' \1 fwhere: 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
( u- y7 G/ x. c# E
+ n0 g+ C& M8 M5 T4 k(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
" h; Z  L! O/ k2 z' f* W! e' v(a+bx+kx)c=sx
$ l. |2 K* ^6 _: U* Cc=sx/(a+bx+kx)

十年移民路_

Solution:

% \0 r8 A" j$ T9 ~+ H) Y! Q7 g( hFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
0 D: R7 O" k1 p7 e& `% qso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
( i/ N" L! ]3 C4 B7 f+ B* [i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
" R" o: J0 J! F8 B+ \5 ]  |! X. @! [* M
: F: P. o; ^& o) u. m, E
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
) p% M& n0 |% T/ ?9 J5 j) j" @which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

{(a+bx)/K} dY(x)/dx=Y(x)
1 X8 }% b$ t' q
from here, we can get:
' v( E+ y) p. ]: G" ~* P
dY(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)
0 Z5 k: D0 d6 \2 z
this means:  Y(x) = (a+bx)^(K/b)
. u& R; R: C1 F9 Z$ Qby using early transform, we can have:

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

6 V3 t  r$ M1 k1 G* z% F! S0 Nfinally:

8 \1 L# A- z0 N' }, FC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)