数学高手请进，多谢！

醉酒当歌 · 发表于 2005-4-3 19:41

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?

多谢了！

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

十年移民路_ · 发表于 2005-4-3 19:52

Originally posted by 醉酒当歌 at 2005-4-3 20:41:  {- K: f" m- D! {# a" Q
请教大家一道微分题，多谢！; B& J# G5 k9 Z5 a1 \: T  t

+ y* g. [% x0 Y0 @4 y8 C2 od [(a+bx) ] /dx = -kc +s
) v+ L8 T; R6 swhere: only x and c are unknown, others are all known,  requiire c = ?. _# ?- ?* |8 _3 p' M7 E/ g- [0 Z* p* k

' T/ W8 V5 @. Y* v9 \# _2 d多谢了！

it is too easy:

d(a+bx)/dx = b

so

b=-kc+s

finally: c= (s-b)/k

醉酒当歌 · 发表于 2005-4-3 19:57

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
' x5 t$ V" |0 Zit is too easy:
) K& E# L, h! Y$ s% s4 i( K, v* [4 |) H$ v+ a) ~" L
d(a+bx)/dx = b7 a: g- B% E6 x/ g% ^" h
6 l* f/ U+ k9 E3 I X
so, j8 G: _3 C5 m1 a3 y6 h2 L
' y' _# U! q, @- l- c8 Q6 ?; l
b=-kc+s; \' L6 t4 s, L/ y. t8 B

2 D8 t& ?# F: s o" b) B$ Afinally: c= (s-b)/k

the right side cannnot contain c, but b contains c, so it is still not solved.

十年移民路_ · 发表于 2005-4-3 20:51

Originally posted by 醉酒当歌 at 2005-4-3 20:41:) Q0 Q6 K5 A: J' k9 ?9 q
请教大家一道微分题，多谢！  l' A' U" S7 e4 G* t. c. f

. A/ X$ X5 Z% ^7 c  T1 M0 ]- td [(a+bx) ] /dx = -kc +s
, Q5 w7 ~8 u! a' y0 S: Qwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?) s  H% L' S2 d. \  S
, t# O* O( \* E, A6 h& ?/ V
多谢了！
$ e: v& `6 R+ G
$ Y. w+ p0 O4 W# i7 s1 o; ^4 k[ Last edited by 醉酒 ...

sorry, did you post another question? your statement is still not clear.

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

please tell me.

醉酒当歌 · 发表于 2005-4-4 11:30

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
6 E$ c4 @( j; y: |. _sorry, did you post another question? your statement is still not clear.* z5 J3 Z) q4 ~/ h, H" @% b$ F5 V5 D

* K, r0 M- J" z5 Q9 O7 h. ca, b, s, k are constants? c means c(x) ?
: \8 N5 G# l: g O8 h/ E" K9 {* {1 u* t) P$ h
please tell me.

对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

多谢

十年移民路_ · 发表于 2005-4-4 18:36

Originally posted by 醉酒当歌 at 2005-4-4 12:30:! F' {5 y7 O9 c6 B
对不起，写错了，忘了变量c,应该是
6 R. O7 z5 f. a# Dd [(a+bx) c] /dx = -kc +s " J% e: Q$ j8 _" v, t7 K: N/ S
_- B# e) K3 M8 U* q/ J$ r
多谢

do you mean it is:

d { (a+bx) C(x)}/dx = -k C(x) + s

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 · 发表于 2005-4-4 18:43

Originally posted by 十年移民路 at 2005-4-4 06:36 PM:3 {5 F2 h4 h5 @- {
do you mean it is:0 v0 k6 G) D# O) @! W  Y
  e, z) C& V2 d' H+ W, _" G( r
d { (a+bx) C(x)}/dx = -k C(x) + s
8 }. f7 h4 v; L4 W; F: @7 X) {+ G6 B! s" r4 O  s
where a, b, s are constants, you want to know what is C(x) ?
* i# u) G( X6 Z! ?! G, y2 o3 ^  T; X- ~9 ^
if 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_ · 发表于 2005-4-4 19:17

answer:

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

十年移民路_ · 发表于 2005-4-4 22:26

this answer is the good one.

procedure:

From:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s

introduce 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:

{(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)

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:

-(k+b)C(x)+s = (a+bx)^(k/b+1)

finally:

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

醉酒当歌 · 发表于 2005-4-5 13:33

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！

十年移民路_ · 发表于 2005-4-5 20:41

Originally posted by 醉酒当歌 at 2005-4-5 14:33:3 R# Y6 @' x$ m; @, b9 M
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。
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 · 发表于 2005-4-6 18:34

9楼的答案有点问题吧。
您说：

d{(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
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。
比如说 dx/dy = (x'y-xy')/y2
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：

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。

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_ · 发表于 2005-4-6 18:53

Originally posted by 龙舞九天 at 2005-4-6 19:34:
3 E% o+ d/ e9 d9 w$ \! I6 B9楼的答案有点问题吧。. V2 N  M2 J/ R; N2 Z6 p
您说：
% }6 m+ X( a, b1 V) b. c0 X4 E& X& S
d{(a+bx)*C(x)}/dx =-k C(x) + s' H! q: w& X; r! }( a. e
so:
8 d; P. M4 O' |" V) k; nbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
3 \# b2 H& R( J. C4 z" L" o% S! ^8 s9 [
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
8 o, O( [5 c# `8 R; x0 Q, x1 e- c但这是不正确的
  o+ H9 D  @! }d{(a+bx)* ...

Please note here:

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_ · 发表于 2005-4-6 18:56

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。

no "x" in that position.

gnome-kde · 发表于 2005-4-6 19:30

d{(a+bx)*C(x)}/dx 意思是说对 (a+bx)*C(x)/x 整个代数式求导数; U- T9 m& _8 A5 i
....0 V- P# `" }3 ?; F! [
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。

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'(x) = [xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2 (应该对吧，我也没验算)

[ Last edited by gnome-kde on 2005-4-6 at 08:31 PM ]

十年移民路_ · 发表于 2005-4-6 19:45

Originally posted by gnome-kde at 2005-4-6 20:30:
9 l) b o+ X$ ^( i这样算混淆了微分和导数的概念。
+ y; Z9 R+ G+ G/ ^3 A5 _
% c o0 W& o+ U. p8 N fd(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. T. D; H R$ H3 c$ a3 M5 t% k

. f* h" y8 I$ @. l. gIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

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

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

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:

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

gnome-kde · 发表于 2005-4-6 19:51

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
" Q4 Q- J8 X% |* t这样算混淆了微分和导数的概念。! ^4 I# T9 z4 v. P

4 B4 b* [; L0 U V1 l3 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
. ^/ M" {: J8 _! `9 v. W+ X* [, j( o5 j% y- H# b
Ah, you made a mistake: it should be:2 s5 a0 G1 f: K% i

3 V! `& D3 F, x [6 e/ m8 sd(f(x)g(x)) ...

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.

十年移民路_ · 发表于 2005-4-6 20:05

Originally posted by gnome-kde at 2005-4-6 20:51:
! y/ Y8 @6 Z9 f. y$ I# V |Yes, there should not be " ' " after "f" and "g", it is a typo. :D
; @5 E. e: r3 c, X, B) x
1 }4 s* N3 |! v8 XBut 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 · 发表于 2005-4-6 22:24

Originally posted by 十年移民路 at 2005-4-5 08:41 PM:( K, l* a$ q; I9 N/ z* O8 V
! ] `" N( ^4 i* S# }# v
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。
. |7 }4 W! d2 D: ?/ K' 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 ...

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_ · 发表于 2005-4-7 20:44

Originally posted by 醉酒当歌 at 2005-4-6 23:24:$ Y7 w4 M W, B6 A* N, y3 H
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。

账号		自动登录	找回密码
密码			注册

数学高手请进，多谢！

浏览过的版块