数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！
3 w  R! r$ C& Z& e0 o3 h
! [- |$ h7 @% b9 H/ ~  {$ sd [(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?
" K. S1 o, I! n: I
4 N; j+ k1 L# w: _( [, `9 I. W多谢了！

+ h/ \) `+ ~5 Q[ 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:
2 J( V2 o6 P7 m( u请教大家一道微分题，多谢！

# I9 }3 |" a& C/ ]# cd [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?
+ j2 a% `$ o- a+ ]
多谢了！

% s2 c( b  V% H7 d3 K  A4 B
it is too easy:
- e# m. I+ d1 b6 S
d(a+bx)/dx = b
; Q/ X4 b' M4 q$ i0 |
so

b=-kc+s
9 N# z7 W6 R2 B1 b6 q3 u
finally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
, z2 n4 P1 @0 J7 Z" N7 z3 n& Dit is too easy:

d(a+bx)/dx = b
# k4 ~, [) d9 L' u: T1 x
so
- t& ~& \: T9 x! I
b=-kc+s

8 S' v7 Q1 X. }9 `finally:  c= (s-b)/k

3 w3 A( d0 x4 M9 J$ F) i' `
$ o$ A# w+ @7 U5 ?the 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！

& k  u1 e7 b1 S  N0 ?d [(a+bx) ] /dx = -kc +s
+ l2 t6 f* T4 L) Q0 J0 O# X& Bwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

$ u0 L& X7 ^3 f5 J" d* E多谢了！
% l0 b& }& G3 J7 z/ M3 N0 x
" B7 ~) ~! z/ N7 @[ Last edited by 醉酒 ...

sorry, did you post another question? your statement is still not clear.
' ^5 a# I  r, o
6 A. t( V7 X7 ~+ F5 q* I" J; wa, b, s, k are  constants? c means c(x) ?
. T# L1 [) P1 u9 R  I# }/ p9 ?+ _
" s8 A2 k, k/ D! o; W& v) }. Eplease 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.
2 D6 K1 [( h) k. U( b5 ~
. @) F! q, x, Ja, b, s, k are  constants? c means c(x) ?
9 I4 n6 s) V6 c; W
+ c! d4 r  b7 r+ S" F; _# Lplease tell me.

# J: B& D2 O; _+ f" u8 D* a; P对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

1 g1 a/ S" K% ]# ^5 Q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
d [(a+bx) c] /dx = -kc +s

多谢

do you mean it is:
  p' O3 `9 A! V' E& _
! o0 Q7 A* `' I! @5 `8 |d { (a+bx) C(x)}/dx = -k C(x) + s

+ s# l) R, C& W, m: {where a, b, s are constants, you want to know what is C(x) ?
4 ~- m- p" R7 i. {
3 |" l- ?* }7 F# M. l( K" s% {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:
0 S' d& x5 ]1 E5 v- zdo you mean it is:
- D' R" ]8 p! Q" h0 u) C
7 _2 [1 G/ ^8 s* _, n- @3 @d { (a+bx) C(x)}/dx = -k C(x) + s

  g  E3 x) ^% @where a, b, s are constants, you want to know what is C(x) ?
. P& n4 m3 ^9 [7 E: Q
. G& F+ n9 E& v1 R/ z/ a3 G4 a1 Jif 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:
3 {7 h: z0 l6 \$ Y# s; ]
, P8 f+ D6 `  q: zC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
" S$ P1 k& W$ d- K; B4 r
) P5 n3 ]! L0 [. _/ h+ x! i
procedure:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
) W8 D- h' Q2 d0 w& G. q
( k+ G) R  ^; ~4 z9 j# z( M3 dbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
3 G4 O' `& y9 A* O8 L
2 |5 g5 b/ M0 p' |, i(a+bx) dC(x)/dx  = -(k+b)C(x) +s
' x7 z& |: D5 }# `: E' j

0 K' U- X2 g1 ]# v% W1 cintroduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
' ~" J3 }7 N# W& }6 b2 D/ n$ owhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
# ^* B2 ]' V/ E1 D8 ftherefore:
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{(a+bx)/K} dY(x)/dx=Y(x)

from 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)
( I* F! D1 R- n: D9 P
" b  k0 ~7 V+ k# U1 I& H% Z. aso that:   ln Y(x) =( K/b) ln(a+bx)
& C2 D- {+ c1 @
+ h" @+ U9 j; u' e8 @, l! Mthis means:  Y(x) = (a+bx)^(K/b)
) u8 p! e& W. W% y- [by using early transform, we can have:
7 [  c% v5 F. b) U9 r
6 i: V0 Z# H, A% N-(k+b)C(x)+s = (a+bx)^(k/b+1)

2 d& O3 |/ E, [/ ~0 A$ `  }( gfinally:
& R; x' r( f# E3 o, G
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:
$ |; y( d2 V& i; M- R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！

  l4 [/ p9 I3 R. Z, w2 ]
3 |+ g$ X( a- \# L4 T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。
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。
您说：

2 m* o* P( x% k  X; id{(a+bx)*C(x)}/dx =-k C(x) + s
4 B8 T( d1 N  O* T; i7 Oso:
5 [# ^* `: w1 m! R( QbC(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
d{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
$ _5 n7 R, L. j- u- t6 D+ A) d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
# w' Z! E# J; Mx' 意思是x的导数，y'意思是y的导数。分母是y的平方。
) c- Q/ Z& g3 K9 `+ l6 v因此：

& [4 V0 ?. ]8 u# V- ]4 L1 Ad{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2
" @2 Z+ U/ I4 a最右边是x的平方。
/ f6 s$ E3 d7 F: m: j
所以您解的有问题。
/ b) F" U1 t4 W; T! `
[ 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。
您说：

. N, i' y  q) e& Qd{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(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
. k2 ?, A6 U. ~, Z' S3 h7 D但这是不正确的
d{(a+bx)* ...

$ v- a! {1 D, c0 r0 ]5 n- i- D/ p! h$ P
Please note here:
$ P9 G' m$ i/ v! W0 b" b
# s; x* f' {7 o: td{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
. n7 F" D2 Z- a6 `9 ]it is a multiple, not division. Right ?

十年移民路_

但这是不正确的
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。
  g3 v1 f6 ^4 e/ V5 @
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
....
5 n* g& T5 z7 o7 w) `' h3 Ld{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

! O) c, a! y/ h( F% S
这样算混淆了微分和导数的概念。
% G9 G# E1 F2 ]6 [, h. j) r
0 K: d0 s, H- {& V7 rd(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

& F2 L" `7 Y) |/ F9 zIf 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)
1 }9 q- S1 C' l) L0 b  N) d
9 Q9 N/ p& f$ f2 c" u2 a[ 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。

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
& q% ~7 o( G% y2 J- q
3 y% F0 a% O4 l! i5 [0 XIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

" p- y8 J. G1 a+ |! I  Z8 Q这样算混淆了微分和导数的概念。
% W6 N$ F9 k: _( N
9 e4 y- c+ S2 H  n, 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

0 k3 s2 G( C( v* L: G, oAh, you made a mistake: it should be:

* D0 C: d8 ^' T) cd(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
4 h# O9 y) i. e: V* g6 y7 Q- r

+ ?; j; {9 h$ {4 Y/ xIf 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)

for h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:

h(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
* p. F6 |  H5 ]& e! ?这样算混淆了微分和导数的概念。
6 {$ C3 W" |' E% X
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
% J% P$ h2 E! Y9 c5 w
, o, |5 Z3 i( R. U0 a/ UAh, you made a mistake: it should be:

d(f(x)g(x)) ...

) p0 n" l1 D2 u4 K- _! M1 P$ {
  S9 b$ ~* ]7 `7 k: rYes, there should not be " ' " after "f" and "g", it is a typo. :D
$ b5 @5 Y+ I& l, {% P% `
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:
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.

7 c8 f! Z' ~7 q% f
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:

! [( T, h3 Z  M4 o( m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。
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 ...

% N0 q. c4 G) N0 a* ]. X
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:
' N& n, ]0 x9 i9 @* y4 m& ~- p0 e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？

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。