数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！

. I7 b* x7 N% Q1 Bd [(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?
, H9 s& g9 b1 D( I+ V
) l9 S# [! `4 V4 W) }: e2 v8 P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:
1 r* x' N4 {! k5 o5 E6 G6 A7 p$ Q5 [请教大家一道微分题，多谢！
/ R) Y& g, F+ f* M
d [(a+bx) ] /dx = -kc +s
+ Z, ]% G- Q/ x& h' Uwhere: only x and c are unknown, others are all known,  requiire c = ?

多谢了！

it is too easy:

d(a+bx)/dx = b
2 u( ^4 e8 X- B$ ]) y7 B
so

  L) @, f2 Q5 M1 wb=-kc+s

finally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
it is too easy:

! n  B3 N# `1 F- Z7 X  {1 _d(a+bx)/dx = b

so
$ m, c# }, O( W% x, b3 j& I2 q3 d7 h
; h% K: ^' ^; f  db=-kc+s
; z) X/ K+ t8 ?
! W  i  j! x/ x; r; Kfinally:  c= (s-b)/k

8 M; F4 P9 d( g: e5 z, O) j3 B" X1 d7 p
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:
9 @2 `7 r* T8 i3 w# G8 L, I8 t- b. G! \请教大家一道微分题，多谢！
9 x9 ]8 m% k7 a5 S# T" @5 i" \
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?
8 v+ U* t) d/ E& t4 N1 v- ~
# e3 M& A( h9 Y% d# d, U多谢了！
& M! y1 t/ o, K! O5 K
2 S6 r# f: y  ?9 @' K# Y4 {! `[ Last edited by 醉酒 ...

' y' Z# O: S; k4 E: |
; A# X& }/ d9 f/ K/ Hsorry, did you post another question? your statement is still not clear.

% F+ s2 H  B; y* c3 m! K# ?a, b, s, k are  constants? c means c(x) ?
! ^" V, |8 w' r7 L
: A0 W/ w  f. x- x. hplease 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.

a, b, s, k are  constants? c means c(x) ?

3 n, _8 P# Z: H7 iplease tell me.

6 O( T9 Q8 l1 A* N" Z" |对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
& |- g; J) H' u( [" ]' ^: t8 ^8 x
0 T5 q6 v% b* N  r. w' k# w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

& T1 ~) v% X; |+ O4 g# m- E多谢

- Q* M( M( w( e$ x! S
  r  y5 Y2 f" edo you mean it is:

0 k7 a: h' D9 bd { (a+bx) C(x)}/dx = -k C(x) + s
/ h  n$ ~; J+ @5 h6 F: B
7 S2 s  S" k- l/ C1 _% g2 {where 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:
- O* F' w$ K' J, C$ a
d { (a+bx) C(x)}/dx = -k C(x) + s
0 v% s+ L; _* T; e# G3 W6 P0 |! u
where a, b, s are constants, you want to know what is C(x) ?
- u: Y3 S8 t) X) C# x: q
3 A. b) ~9 P' M" L' k1 L  {if so, this needs to solve the differential equation. please comfirm it fi ...

9 U+ l$ Z5 S3 O! ]
2 K) h' X7 e% K( v; W, P$ T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:
) Z! f7 Y' G/ M
7 i$ P9 M, F$ s! K) KC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.
4 S1 Z8 d8 e( a* C
5 K. G7 s. G8 X5 I8 Y
procedure:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
! w. p# a3 {/ w0 z& b1 h' v: xso:

) g) {# b" o+ T& f1 u& nbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
; s' a6 s; [8 i) ri.e.
8 e( G/ R6 J! Y( _6 P
) b: f6 v* L% |5 }) e5 D9 s* ?(a+bx) dC(x)/dx  = -(k+b)C(x) +s
' ?+ O* Z7 Y7 L" T

- X* c. ^0 s3 s: d+ \$ vintroduce 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:
! L9 N. J4 {- m1 K
2 G; l0 `$ E- L+ J2 M9 \{(a+bx)/K} dY(x)/dx=Y(x)
; N$ t# |  `: Y% F: _
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)

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:

2 q) m* e5 h) i-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:
; ^- [: N( a/ X! z7 k- m5 k, {
1 ~; W$ P- u3 r) ?% |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:
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！

" y2 d' W9 A$ }( i" g" ^  s7 D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：

d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
# y8 `1 H2 q3 d- l: @) ubC(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
; N: P% o' p# s8 Y( w并不是   (a+bx)*C(x)   的导数除以   x   的导数。
7 K6 `" _+ ^1 b( T  y7 S比如说   dx/dy = (x'y-xy')/y2
1 W& u6 i; u$ {6 O5 x  {+ w2 t4 u  {x' 意思是x的导数，y'意思是y的导数。分母是y的平方。
因此：

! o, e* m; t* zd{(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。
1 |5 t& j$ G0 s
0 V" j: s' o5 F" c所以您解的有问题。
/ `+ Y7 ?. p1 P5 W4 l
0 T3 c5 L( B' x$ U9 J8 q/ ~[ 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：
( \- Z9 N7 V! B+ W5 X
: {% p% T0 E) u8 V" Td{(a+bx)*C(x)}/dx =-k C(x) + s
9 O; b# Z: o0 {so:
0 {+ M: o; s/ @2 x* tbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

7 W: Q4 p. K  P- G" \7 E* w! c9 E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
8 f; K7 p) J$ z) C2 |d{(a+bx)* ...

: M' |1 r5 X+ f1 a, g) U
4 \) |9 [! _+ M& Y. j& ~Please note here:
$ R" A( C$ b) ~
d{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
: F# M( o% W  c9 Q+ s: b  Rit 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。
; z5 D, D. S' d9 h9 H- Y
$ c# u! ^& q( H$ a% u2 A, hno "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
....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

这样算混淆了微分和导数的概念。
  @9 w# b& ?) M$ S
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
  F' v8 _" h  k  |# T# I
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 b4 b1 o8 s& y2 `. E% `1 ^; }[ 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:
, a, E! M! X, |5 {. M4 a0 _" h: C这样算混淆了微分和导数的概念。
& K" R4 Q8 w& Z4 Q5 a  N
1 f9 u. W7 b0 H# |" I  id(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
, e$ @) G# r3 O6 [' P: U
) y/ g4 R1 @% J* A3 r) XIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

! ~4 D/ q$ e- A7 I
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
" h6 A: g7 _- q
Ah, you made a mistake: it should be:
% E) m9 b% N% D( o8 U3 Q
d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx

# i- O" b; t$ S: G9 N
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)

for h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
( @* T- ^7 @8 f# ?
/ n) b" d- M2 }) l" fh(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
这样算混淆了微分和导数的概念。
3 @/ d# Y. o$ j6 b' p) B  z
9 P, I( @' g$ R  ?. vd(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
% X, h7 k' O* ~. y
2 ~% _0 l& j+ X2 TAh, you made a mistake: it should be:

' r3 _, o3 L; N% t/ S; wd(f(x)g(x)) ...

8 A4 S$ r1 H  ]7 M) w
# |( b# r4 t" F! y" W/ vYes, 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:
; d' A# U/ C- C8 X( _/ m) CYes, there should not be " ' " after "f" and "g", it is a typo. :D
- n% e, w& `/ u( v7 }8 Q1 k5 D
( K3 O: R+ ~1 J6 _: c" rBut for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

: M# j- Q( c/ y8 q  f$ g
$ g5 z0 A3 ~. [' ]& P5 kThanks, 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:
- n1 I! [% d8 T9 V' D2 c/ d
5 b. h  o2 t$ S1 ?$ g$ a0 u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 ...

& m6 M2 J/ e% v! W* S4 T$ R
  h* o/ |3 C4 p8 t; [8 g* k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？

4 T$ N/ k# `# b5 w7 G- l+ V- y5 }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。