数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
' E- M/ ~+ ]& Y; t
d [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

9 f6 E: S7 D3 b! j0 x# q7 n0 Q( r多谢了！

[ Last edited by 醉酒当歌 on 2005-4-3 at 07:56 PM ]

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-3 20:41:
请教大家一道微分题，多谢！

4 a( m* G7 g+ l/ P0 K3 u& Vd [(a+bx) ] /dx = -kc +s
0 @; B' M: \# q" ?8 M4 ^8 wwhere: only x and c are unknown, others are all known,  requiire c = ?
9 W  X/ z  E# `% w1 B6 U# @7 h
多谢了！

" N$ c  ^! W0 o/ b( F+ y' e' @& D3 u
it is too easy:

d(a+bx)/dx = b

( g7 m& ~: C  m) e  Kso
9 v4 ?; \1 |3 ~: N' T; l, A
b=-kc+s
9 `0 x5 Z2 F8 q% J# y+ b
. q' ^2 Y% Q& P; y, |0 qfinally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
  f- P) [1 ]. K' |it is too easy:
3 J6 z" y! ]# w! \9 U
d(a+bx)/dx = b

so
- \3 a6 t) m# m
b=-kc+s

finally:  c= (s-b)/k

) I9 M) u$ Y2 w2 Z
) ]. m3 C6 b3 S% h# Qthe right side cannnot contain c, but b contains c, so it is still not solved.

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-3 20:41:
请教大家一道微分题，多谢！

d [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
4 \; i8 Y  Z6 `4 f/ M) L
; e1 f; p% v0 ^# k2 q8 Q多谢了！
5 S9 E4 h6 `. g/ K. M. u
[ Last edited by 醉酒 ...

, m! a8 n; ?% z9 Q; Z1 `! C+ ~sorry, did you post another question? your statement is still not clear.

a, b, s, k are  constants? c means c(x) ?
) x# W+ v, M& Z5 _  {8 K
+ L; a" _. J/ }5 N8 w. wplease tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
sorry, did you post another question? your statement is still not clear.

0 x  ]" [( A5 ?( V1 y2 z/ ca, b, s, k are  constants? c means c(x) ?
$ g$ z+ \, t9 N/ \, u4 n9 [( ]1 G1 l
" f3 x- J) @* N( Y  uplease tell me.

: V: t) H1 Y* J' [( t对不起，写错了，忘了变量c,应该是
# }' Z7 O/ F7 gd [(a+bx) c] /dx = -kc +s
+ L% U1 m, z+ C
( W& K7 Y* g' O# h多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
对不起，写错了，忘了变量c,应该是
; N8 X' J2 y2 F% Ad [(a+bx) c] /dx = -kc +s

多谢

% \' W+ c# o9 D/ A- G* }' i6 S
do you mean it is:
5 {% p5 {/ k' t5 m3 I+ _
d { (a+bx) C(x)}/dx = -k C(x) + s

5 b* [% f6 M. w( x. x  K2 D, O7 Twhere a, b, s are constants, you want to know what is C(x) ?

if so, this needs to solve the differential equation. please comfirm it first.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-4 06:36 PM:
do you mean it is:
0 a1 ]( \' N2 V  V
d { (a+bx) C(x)}/dx = -k C(x) + s

; w& n$ w! m: v( v7 z3 nwhere a, b, s are constants, you want to know what is C(x) ?
, ^) R3 q2 F  I% P( a
6 q/ a1 h7 D& r2 P0 Cif so, this needs to solve the differential equation. please comfirm it fi ...

多谢你,你说对了,就是说用x的函数表示c. 也就是 c(x), 多谢你了。请帮忙。

十年移民路_

answer:
( C) _+ Q4 Q: p" R9 @3 G
C(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
/ _+ d( w/ d$ [' R" [0 `' z9 @

: c1 `7 k; n. Sprocedure:

0 V" ?6 O" R0 N2 {2 }From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
% C; L: A% u7 q
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
$ j8 @. Y5 J. k8 F2 B( u
8 Y* G: I  C6 w* C(a+bx) dC(x)/dx  = -(k+b)C(x) +s
; f8 r# C" x- O( `4 `
/ b, N5 t- o* h
introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
% P" {6 Q% W+ i, ^6 owhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

# ]: I7 J3 X  m& y3 f7 C+ t3 @{(a+bx)/K} dY(x)/dx=Y(x)
  p) x: A0 M5 _+ E5 q1 W
+ k6 [* R$ S, z% Z+ zfrom 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)
, q. a0 h  d, K9 k( A9 ~6 w
so that:   ln Y(x) =( K/b) ln(a+bx)
) u# C' m' |/ o  h9 ?' h6 N9 O
this means:  Y(x) = (a+bx)^(K/b)
( Z( b* u, L+ C/ s5 V% fby using early transform, we can have:

0 Q1 p  @: l2 n9 l) F, q-(k+b)C(x)+s = (a+bx)^(k/b+1)
* W% {: R) Q% d; v; A+ v
- a+ H+ Z4 ~& d3 `& K! ~finally:

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

醉酒当歌

果然是高手，多谢了，打印出来好好研究一下。高等数学学得实在不好。你一定是工程师吧，而且数学相当好。再谢！

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-5 14:33:
5 v% x6 ?1 h5 B% K7 W果然是高手，多谢了，打印出来好好研究一下。高等数学学得实在不好。你一定是工程师吧，而且数学相当好。再谢！

请你再仔细检查一遍小地方，但我可以保证方法和过程是没有问题的。
" m/ C+ I& j/ l' d是的，我现在是注册工程师，在上大学时，曾把“基米多维奇”和“樊映川”两本习题集通做一遍。（共大约一万多题吧）。这种题真的不是难题。

龙舞九天

9楼的答案有点问题吧。
' u- x% d' _, r, z/ N您说：

2 L3 Q+ y. g8 J( L, P$ g7 Z: q4 n# l- gd{(a+bx)*C(x)}/dx =-k C(x) + s
" ~' u6 n( `% r% Fso:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

" T: c4 {/ L; z2 R5 V( E0 }' b0 K也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
但这是不正确的
1 R9 h) c$ C+ Z) F7 Z: X5 h6 U& ^d{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
并不是   (a+bx)*C(x)   的导数除以   x   的导数。
比如说   dx/dy = (x'y-xy')/y2
5 Z2 A  S1 b) T  K- R/ s4 z& O; l! Jx' 意思是x的导数，y'意思是y的导数。分母是y的平方。
因此：

d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
最右边是x的平方。
6 |9 ^6 V) [! `8 z5 f
所以您解的有问题。

[ Last edited by 龙舞九天 on 2005-4-6 at 07:35 PM ]

十年移民路_

Originally posted by 龙舞九天 at 2005-4-6 19:34:
9楼的答案有点问题吧。
& X6 L% E* R5 `* L3 m. \您说：
: e: |+ B. m$ `$ M4 i" B
  K: b: t$ u( j( Wd{(a+bx)*C(x)}/dx =-k C(x) + s
% V3 G  g4 \% J& mso:
2 \' ~5 r4 G% h. GbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
但这是不正确的
; G- g: m4 F+ s, Bd{(a+bx)* ...

Please note here:
5 q+ o$ n! c$ F8 [' ~  Z( H! X* X
d{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
it is a multiple, not division. Right ?

十年移民路_

但这是不正确的
6 }+ k, G2 t' x( ?' B# \d{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
并不是   (a+bx)*C(x)   的导数除以   x   的导数。

no "x" in that position.

gnome-kde

d{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
/ E1 _$ p, Q( x( h6 s; G4 k) D8 Z  O....
! E0 v* i6 ~4 q. fd{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

0 M6 L- r( v3 i( H. C( J4 e
8 M7 j1 s* n* \) {% e这样算混淆了微分和导数的概念。

( q/ v6 E$ N6 sd(f(x))/dx = f'(x), d(f(x)g(x))/dx = f'(x) d(g(x))/dx + g'(x) d(f(x))/dx 这是基本的微分四则运算
$ B5 ?9 H, l$ v+ l* {
If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h'(x) = [xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2 (应该对吧， 我也没验算)
& |( _0 P) n; }0 t
[ Last edited by gnome-kde on 2005-4-6 at 08:31 PM ]

十年移民路_

Originally posted by gnome-kde at 2005-4-6 20:30:
这样算混淆了微分和导数的概念。
( Z! [0 Y5 t3 l
d(f(x))/dx = f'(x), d(f(x)g(x))/dx = f'(x) d(g(x))/dx + g'(x) d(f(x))/dx 这是基本的微分四则运算

If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

/ d& g4 }3 }! X9 C" M9 R这样算混淆了微分和导数的概念。

: R/ \  ?. S* w/ U. bd(f(x))/dx = f'(x), d(f(x)g(x))/dx = f'(x) d(g(x))/dx + g'(x) d(f(x))/dx 这是基本的微分四则运算

Ah, you made a mistake: it should be:

d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx

" h0 J) Q' |+ S
) e3 T3 Q+ F9 i( ]2 {% ?2 DIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h'(x) = [xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2 (应该对吧， 我也没验算)
" J" q& Y  ]( n
for h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
1 e3 L. |) {& ]/ ~3 r# ~0 ?, W
% ~  u! d  m# y) B9 @* D5 Y! ph(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
1 g( d& U' Y  n; @; G9 q这样算混淆了微分和导数的概念。
. ?8 T8 [# C( D0 f9 d$ j& u2 I
d(f(x))/dx = f'(x), d(f(x)g(x))/dx = f'(x) d(g(x))/dx + g'(x) d(f(x))/dx 这是基本的微分四则运算

2 N" C# n3 x9 L* m7 {2 z5 ~: SAh, you made a mistake: it should be:
- }  |, I, {1 o4 y
* ^; k+ i5 }1 v  w7 Ad(f(x)g(x)) ...

: Z! k2 g9 v! B9 P9 A  \) GYes, there should not be " ' " after "f" and "g", it is a typo. :D

But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

十年移民路_

Originally posted by gnome-kde at 2005-4-6 20:51:
1 K' y/ R: G$ [1 `" `! H  g9 D2 }Yes, there should not be " ' " after "f" and "g", it is a typo. :D

But for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

Thanks, I am happy here to discuss math.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-5 08:41 PM:

) D! e( {0 \5 u6 }; l, V& n# H1 Y" g请你再仔细检查一遍小地方，但我可以保证方法和过程是没有问题的。
0 M- r/ G2 \" ?! G是的，我现在是注册工程师，在上大学时，曾把“基米多维奇”和“樊映川”两本习题集通做一遍。（共大约一万多题吧）。这种题真的 ...

; d- W+ C7 U+ ?1 q- @佩服！是不是微分方程如果不能transform就不能求解了，有没有现成的微分方程求解公式？

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-6 23:24:
佩服！是不是微分方程如果不能transform就不能求解了，有没有现成的微分方程求解公式？

1 _# N$ R# O3 _! r8 p. S2 X
理论上，只有很少的几类”齐次常微分方程“可以写出其”封闭的解析解“。不作变换，微分方程是非齐次的，可能不太容易作到。一般地说，没有什么现成的微分方程求解公式。但对”线性定常齐次微分方程“是有一个较方便的解法。