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！
/ N; L) {( v( s% w  C9 Q- ^: ?) }" A
. Z( w5 ~5 b0 |5 T8 T! ~[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

(a+bx)c  = (-kc +s)x
$ f, G* [) t/ z: Y; t. aac+bxc=-kxc+sx
9 j) {& i0 C3 Q  a# [(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
3 e2 t5 a3 N4 r, D! z+ Z) h
0 ^5 [- G! \2 ObC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
& N+ \9 N6 V) P
/ {! y! s3 X! ]0 K) w; b4 a4 O(a+bx) dC(x)/dx  = -(k+b)C(x) +s
! Y. ~4 s: N( w, Z$ n+ i+ A

( S( s% u3 R; ~9 n' mintroduce 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:
" |6 F1 Q0 K, ^! c$ k6 l8 g4 {
, D5 G( w1 d7 L8 F! I{(a+bx)/K} dY(x)/dx=Y(x)
# B9 F" p8 w# ?( i. O! _
from here, we can get:

! Z0 r! ^3 y( ^, m# s4 o- CdY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
0 W9 k) c. \# J& `4 {
so that:   ln Y(x) =( K/b) ln(a+bx)

  O, x8 @3 g! i3 ~9 J) j0 j4 S  ^) Othis means:  Y(x) = (a+bx)^(K/b)
! V4 X) M1 R3 K4 }9 t- _+ H* pby using early transform, we can have:

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

6 {1 H2 d0 O" u( Y+ afinally:

* ?+ y6 \2 d% vC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)