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?

- O" J9 D6 m8 J' H多谢了！

: O! Z0 A3 g3 e[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

4 U$ z. I7 t2 f. e9 C(a+bx)c  = (-kc +s)x
1 J; |, a6 ^$ C$ v: @/ cac+bxc=-kxc+sx
(a+bx+kx)c=sx
" U0 r, _# d5 e' m& ec=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
" H+ r2 b- m* X9 ^' U$ q2 yso:
. V  t( r: g1 ]( `3 [
4 U- l/ B4 q4 A! `% F1 PbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
* V2 u  p, b  l( v' Q, T- h2 F
; \* M* ^+ I2 Z# j0 I0 ~/ B* G(a+bx) dC(x)/dx  = -(k+b)C(x) +s
- V# K: k0 d2 i+ s- r; i

introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
which means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
9 D& S1 \/ B: c7 @9 P9 ~2 X. N
0 P# E0 j. J) r7 W{(a+bx)/K} dY(x)/dx=Y(x)

, Y* {/ R; x( o9 E2 zfrom here, we can get:

0 h1 ~- ?- y$ R% k. X* \/ K$ f1 `8 U- JdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
. y, Z' N" [4 k
so that:   ln Y(x) =( K/b) ln(a+bx)

6 \+ d2 \# E; Tthis means:  Y(x) = (a+bx)^(K/b)
9 v! G! z' o6 k8 x7 u1 d7 F% Sby using early transform, we can have:
6 G& x  n% D' g  M
& b% ]  j+ y0 J8 _! d1 B) H-(k+b)C(x)+s = (a+bx)^(k/b+1)
6 W, R! ?, i; }* d( u8 @  B* f3 o3 @
finally:
, V+ o# V1 e3 l1 ?/ Q
C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)