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）
5 A1 \7 z" P8 I
& ~9 p( S7 z& s% S  ^2 m; V8 i, bd [(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?

9 I, p& S8 z+ N+ l% U$ w多谢了！
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) G, d7 i9 Y6 F( B[ 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:

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

3 F1 R- ?2 a5 G" ]& W" z/ PbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
: n, }; l& A9 D7 x7 R5 ji.e.
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(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
therefore:

{(a+bx)/K} dY(x)/dx=Y(x)
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7 H( j+ V0 e6 E; Yfrom here, we can get:
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dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = K/b {(a+bx)}d(a+bx)
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" P. H! _' N" n2 p8 Z- C3 [! mso that:   ln Y(x) =( K/b) ln(a+bx)

2 n6 _+ w6 z0 Jthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
  h/ f8 ?9 R$ @5 b
9 T6 F6 m: ~' e( i6 I1 ^% j: b-(k+b)C(x)+s = (a+bx)^(k/b+1)
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3 |; ]+ F' V+ ^1 O, W# yfinally:
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0 {2 a7 w; O! h) A2 i  ?C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)