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）
$ f+ R; V) P$ L1 @9 R! \
d [(a+bx)c ] /dx = -kc +s
4 @9 o! h+ X; _, _where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
  M8 O. B* a6 j, K6 k: B8 ^
多谢了！
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5 d& M/ ?: K/ ]: H[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

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

! b* ^0 n, [6 h. R5 z6 a) h$ M(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
3 f( v; Q/ v+ i, vc=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
0 r* r. r5 K# i" Y7 ^6 L" ~3 |
) k; w( j" U! f7 U/ C. ?1 E3 Y5 p(a+bx) dC(x)/dx  = -(k+b)C(x) +s
! K% E' x" o- N9 {
) O4 I& f* n' K( C( s9 q
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:

5 h! ]( b9 K* d- L$ D{(a+bx)/K} dY(x)/dx=Y(x)

6 u# ]7 D1 o# o8 D; B; {3 nfrom here, we can get:
% k6 Q) w& c6 n1 Z) B6 L
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)

$ S* S3 h+ Q3 kso that:   ln Y(x) =( K/b) ln(a+bx)

2 {3 B: i; m' B, i6 Y( X+ [: dthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
0 S8 Q/ ~' _( \% N3 w/ }
8 I* E% F1 c( c5 q6 B3 o7 {* O3 L, v-(k+b)C(x)+s = (a+bx)^(k/b+1)

, T; j$ a0 L2 t. y8 _! Y, Zfinally:
* m# x8 u  b) ~+ ^% y. F
, h. |* T  a: M5 H/ L6 CC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)