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?
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多谢了！
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0 J5 t1 u" y+ ^8 ~2 ^" v2 {8 l[ Last edited by 醉酒当歌 on 2005-4-4 at 11:33 AM ]

sindowa

d [(a+bx)c ] /dx = -kc +s
& ~+ Q  c" s2 F1 L  U8 r- E+ w1 U
(a+bx)c  = (-kc +s)x
ac+bxc=-kxc+sx
3 \+ s. x+ v9 ]# a  u& R(a+bx+kx)c=sx
, q) s  r, S4 e, h' l  Uc=sx/(a+bx+kx)

十年移民路_

Solution:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
0 [# _! U7 k0 k/ Pso:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
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(a+bx) dC(x)/dx  = -(k+b)C(x) +s
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& Y! a$ z$ f. C8 Q% ^* X# U
2 c/ J" R5 M' gintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
% W# ]. F+ A2 H, S0 i. @3 U' dwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

{(a+bx)/K} dY(x)/dx=Y(x)

8 Z; U, D* o. |) g6 t6 q- s% n3 J9 S' Mfrom here, we can get:

, U1 a. M, a9 v. ]! d- j  G/ ?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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so that:   ln Y(x) =( K/b) ln(a+bx)
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this means:  Y(x) = (a+bx)^(K/b)
0 W5 \1 r: _0 X5 H0 s) E/ Dby using early transform, we can have:

) j% A" _1 H5 x  f# S$ e-(k+b)C(x)+s = (a+bx)^(k/b+1)
% B0 p$ L/ |0 c) b- s; }
finally:
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% R9 C& t) ~) b. K. V: i/ RC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)