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）

4 b" J/ \0 A2 G) l7 Z0 Y0 Ud [(a+bx)c ] /dx = -kc +s
3 E- U/ ~; h! x- @3 m" }7 Lwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

' p: B& H9 G9 b: r" `# v1 ?多谢了！
$ }' p4 P" }9 ?4 k. u% P7 r
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
# M) R/ [2 W3 l( s# V0 K
(a+bx)c  = (-kc +s)x
: H3 c- S, k, T0 ~ac+bxc=-kxc+sx
(a+bx+kx)c=sx
4 r* l4 ]2 x, [: }" o+ @% W" Yc=sx/(a+bx+kx)

十年移民路_

Solution:
8 j+ ?3 X% M. w* k" M. }0 F
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
1 d$ s' K5 c, y9 t. Hso:
2 R+ K! h. Q8 E; d3 j
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
! D/ N7 D  a+ i$ q9 ~3 \
(a+bx) dC(x)/dx  = -(k+b)C(x) +s
8 V' ]; D6 _9 u

- ~# j, Y1 F2 eintroduce 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:

/ e  Q: @4 n5 ?0 R6 l' e{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:

dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

5 b2 B* i- I" @8 Sso that:   ln Y(x) =( K/b) ln(a+bx)

this means:  Y(x) = (a+bx)^(K/b)
' M3 q. L# T  _9 h5 rby using early transform, we can have:
; o5 J5 R" ?3 E0 n2 T. V
8 z  F. i, z( w" R* i-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

$ t/ v- ], R/ [C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)