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）

+ w5 t- z  e  Id [(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?

: t) v% L1 i# s$ p多谢了！

[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
6 s) `+ N2 L& ]4 y8 L1 \* W3 E
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
# Y9 I9 i. O% a$ {(a+bx+kx)c=sx
  D5 F- [- f2 d' b6 E# U# Lc=sx/(a+bx+kx)

十年移民路_

Solution:
- y) x+ ^3 `6 z! h, g
From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

0 l! u, D( }! EbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
0 L) r/ z/ ?/ c, w& G7 ri.e.
. k. J  \7 r; M6 B2 ]) I
! A% u0 p; v0 C+ D8 a(a+bx) dC(x)/dx  = -(k+b)C(x) +s


4 @$ W6 B) w0 E( A2 R) |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:

% j4 ]5 x  Q. l& }# m5 K  b1 b{(a+bx)/K} dY(x)/dx=Y(x)

2 Y' X6 b  t0 n) Ofrom here, we can get:
6 ?9 T8 K* n3 H% L
dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
: n5 M% K" Y$ j3 ?
so that:   ln Y(x) =( K/b) ln(a+bx)
! I/ S7 }5 S9 [7 b* E/ b) a
this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)
6 T4 q& T9 f1 K! y
finally:

C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)