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:
5 ]+ r3 S# B8 \9 `9 f3 R: i$ S请教大家一道微分题，多谢！/ O0 s, [7 @0 e5 r. I+ I
n9 D; y- c4 ~# {0 H( V
d [(a+bx) ] /dx = -kc +s * P7 Q# G& v% v5 d2 [7 `. J
where: only x and c are unknown, others are all known, requiire c = ?/ c) N- Q: p0 y f

+ x5 T$ z5 S% m7 V7 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:
  p- w& s$ e9 O, b$ \1 ?/ git is too easy:$ a/ d/ r( _+ ^" L

1 i+ b- j# }) v4 f$ d8 ]2 Nd(a+bx)/dx = b# y( R* d5 L/ F
4 Q4 e- `7 J4 Z' g/ D5 t
so9 Y2 V  R% C: d5 v
+ g$ P8 |5 h# {# n; j+ B6 q
b=-kc+s
$ o  x' `& v1 x& z* ?- J$ Z0 n' l! I* y: ?3 J! @+ F8 W9 |, E
finally:  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:' k/ R3 |$ c/ N, \3 Y
请教大家一道微分题，多谢！# S! _3 \; i( x: t2 J
% [8 [. |4 t8 E! n) ?! y; x
d [(a+bx) ] /dx = -kc +s
4 y6 n/ [9 }7 s/ w' p0 \, Z: [/ rwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
- C1 l% A( i& B4 |
1 J% b6 `* v4 d$ }多谢了！
/ i. L! m) M& Q; w( m( O% F9 V- I% s( O2 S" p+ T
[ 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:
/ n" B+ M8 |. K# G. zsorry, did you post another question? your statement is still not clear.- T" C- s/ }. Q* p. n6 D

a, b, s, k are constants? c means c(x) ?( O- N( ?) r8 c0 X- @) @* ?

L( f* E. t9 H1 O' Z9 C3 `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:
6 _6 b7 L$ ~$ b1 O- _) S6 A对不起，写错了，忘了变量c,应该是
% S7 ~# I) ^8 vd [(a+bx) c] /dx = -kc +s ) A j) C7 q. n: |: B# `* P8 d5 k
$ p3 o% k" y( c$ H5 [# H5 U
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:
: r- L# [* M" g+ [do you mean it is:
' x3 g& Q" x8 f3 ]6 X9 l+ x# r0 _2 h' I2 |+ b* s% P8 D3 h$ `
d { (a+bx) C(x)}/dx = -k C(x) + s% ^0 C& K) S) h2 U7 T

6 E/ O5 M! i% l# X3 v4 E4 u* ^' Vwhere a, b, s are constants, you want to know what is C(x) ?+ o3 x6 M. Q( i

! S g% o/ H/ zif 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:
4 z# w1 K3 G$ ~) w j; D9 f: F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 x4 V- [( A, D' n2 u
9楼的答案有点问题吧。
6 c! a9 @2 L/ y7 K8 U3 ~您说：7 }' r+ o. L: J! e3 k

# U" f* _+ l5 @d{(a+bx)*C(x)}/dx =-k C(x) + s5 v" @6 w- }2 S7 A! X* B: S
so:
1 p! F" ?" f( j4 T! u6 ObC(x) + (a+bx) dC(x)/dx = -kC(x) +s
7 w) N0 i2 f' j# I+ S1 y/ n7 [3 U* A& i, J5 F7 j9 q
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx$ X: ]2 ~' B( X5 A
但这是不正确的' n3 I7 h- N5 `6 L( i
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; k$ H- D4 T; ]: {
....7 x! B! ^6 E+ l8 F% O
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:
8 L2 c! b" H8 k- \9 Y这样算混淆了微分和导数的概念。6 d0 P- h* M- c0 Z/ |
* _; e9 D$ B8 o
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
8 x% _, R; i) U/ t9 v" L0 y/ P6 E' y3 k4 \4 \0 v
If 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:0 f" u1 B" B5 L  ?/ l
这样算混淆了微分和导数的概念。3 \6 ^  n0 X- j/ S
5 l+ c$ j% N9 e0 G# ?6 Y
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' m; A/ ]$ W& n" K% ^- ?
1 ]& Z$ _* y! t0 @
Ah, you made a mistake: it should be:
' o! [$ x  r/ W6 v
! M6 r6 P: f9 f: ?9 j3 ^d(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 20:51:4 C" }- s( z) r5 R/ W( D* ~
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
7 h# i4 R5 Q6 J/ q8 w; Z9 `8 K) ?7 V" l+ v
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 · 发表于 2005-4-6 22:24

Originally posted by 十年移民路 at 2005-4-5 08:41 PM:
$ X9 X3 r5 z5 |0 O( y) G, Q+ a4 _# _3 P! L: v% `8 j3 X6 i
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。
: T9 A1 e5 u1 {7 w1 F# c! i& G v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:
0 e$ ?2 O) f9 L/ \# b& ~' R$ s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。

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