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）
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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！

7 x# z9 L! {0 }3 D& V# A  `[ 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
ac+bxc=-kxc+sx
(a+bx+kx)c=sx
c=sx/(a+bx+kx)

十年移民路_

Solution:
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- ?0 G8 N, P/ s1 e' e( PFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
& |3 h! U; }7 Q/ bso:

% u( W% W' x$ g! y& ibC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
5 I- s! o6 l& u+ f, n+ B% V4 X( X
(a+bx) dC(x)/dx  = -(k+b)C(x) +s


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
. e8 C$ h9 _* x0 q0 Q6 s; d2 Wtherefore:
% |; T) n) J' w" g) j+ k" H9 F9 I
{(a+bx)/K} dY(x)/dx=Y(x)
9 W- f! }$ `* u- Q  Y0 {1 m% M
6 O- F' m0 j, Y0 }  V9 yfrom 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)

so that:   ln Y(x) =( K/b) ln(a+bx)
& a. T( P! ]2 X: Y# H% m! i  s
# O" b  N! Y( x, Hthis means:  Y(x) = (a+bx)^(K/b)
& _/ G" E7 T* ?by using early transform, we can have:

-(k+b)C(x)+s = (a+bx)^(k/b+1)
& s+ a; Y7 l2 k/ m
  ?& G: S6 ~% y2 n, O0 t" xfinally:
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C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)