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）

# t$ o* c* [# c1 Nd [(a+bx)c ] /dx = -kc +s
4 r6 D* l( L, q9 vwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
' k; e+ k4 X$ S5 V$ Y8 Y/ R4 t
, i8 d, Y- l% m多谢了！
6 k4 Q- o7 z; y3 ~8 H
[ 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
7 O. Q4 H: l2 Pac+bxc=-kxc+sx
(a+bx+kx)c=sx
+ s$ a5 I9 n3 Z9 H* n/ }c=sx/(a+bx+kx)

十年移民路_

Solution:

- A6 G% ~3 v& g$ N+ b/ m7 r( A( pFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
4 [  t& `' [' ~
3 K9 O6 d, X) A4 a; W" V(a+bx) dC(x)/dx  = -(k+b)C(x) +s

- e0 O# a1 s( C
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
+ }. v' E& v2 _( Zwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

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

& }! k9 d( _2 {# U% w+ x! \from here, we can get:

; v4 {+ ]! M" |+ C2 T9 UdY(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)

this means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:
) p6 W! |9 ]/ j, m$ w6 q
# I" r. n+ A9 U' g6 j-(k+b)C(x)+s = (a+bx)^(k/b+1)

( M5 F6 @7 |& ^! Xfinally:
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C(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)