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）
2 U2 y) m) C0 L5 O+ S
d [(a+bx)c ] /dx = -kc +s
" H; I5 m# S, O$ {where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
! |- u2 y2 B/ t2 w  v
多谢了！
, [5 c6 v1 {& P$ [. Q7 |  T* P
[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

2 e( `' A! s: {# Z" N6 j( |(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
8 w! ?. e9 ^! C* U) Y4 S' [: I(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
) x$ Q" j; M  V* [; u; v- Gso:
* F% h8 o, J  I+ [
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s

# t% z1 c  x4 `" p0 u; ?
0 p5 \6 U% v# U$ s9 v. uintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
3 `* ]& [8 v* ywhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:
4 V9 r- u+ F1 p. i0 w% f" W# Z
{(a+bx)/K} dY(x)/dx=Y(x)
' k/ \# \5 N" q( S  f4 l) [
from here, we can get:
4 v1 E) Q. p: Y) h' w3 u0 e4 l% }
, x8 l: r; m' @  j  _- ydY(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)

/ A( x" L2 A/ x" t# R+ Nthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

# T5 ^/ B4 ?/ q-(k+b)C(x)+s = (a+bx)^(k/b+1)
& B" ^" m9 R/ {/ C/ ~) W
finally:

0 N! m) S$ P/ tC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)