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:
  ]& [6 V- f) v" m+ W5 y! ]9 J" Q请教大家一道微分题，多谢！
3 S+ V. ^! |' M3 Q
$ r1 y7 ]4 w- g! m$ \9 g2 r' pd [(a+bx) ] /dx = -kc +s
3 C7 |" Y# d. |1 B0 C  d8 Owhere: only x and c are unknown, others are all known,  requiire c = ?( D5 K; _/ e% }% W- a' G" f4 {

4 [' ?( i6 _3 J9 C  o# o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:  v- P9 l, h! z1 p! v( I
it is too easy:  C. i$ L+ Z/ ~9 p# ~1 k

7 h, c9 d: `2 Id(a+bx)/dx = b
6 `! K' p" r, g& {) _" c: p/ \9 g0 |* }5 Z
so4 |9 m$ g* I& Y% n, A* @7 T

; H" }4 t) X5 U& \$ _% M( ?b=-kc+s! e" E+ f: O% [) o8 A8 P
% ~1 s+ f! S0 H" a7 F# `2 v2 t, C5 T' O
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:
% V& A$ G2 d+ ~请教大家一道微分题，多谢！
6 W. `+ T& t! c9 i0 ~8 \- ?# G- @& K6 S% u
d [(a+bx) ] /dx = -kc +s 7 m# r1 [! h' k1 [$ N  t& I
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
- D: a/ u; w  }: Q& T( J
1 L, H% K) `+ S多谢了！( u7 E8 y$ U9 R0 t9 t( h

) s2 A0 U- S0 L6 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:4 E- X. E% M: c; W, f7 I
sorry, did you post another question? your statement is still not clear.
/ x8 T; U6 G7 I- r& i3 \" g# H3 b& H1 `6 J$ x, z1 f% }
a, b, s, k are constants? c means c(x) ?
" @& ~3 D2 D3 X* H) w7 I- a1 z7 Y+ l7 C6 M! ~
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:
5 Y% {9 `4 @* v4 {7 b: `对不起，写错了，忘了变量c,应该是
6 D) F- N4 F; P8 p: a0 U( fd [(a+bx) c] /dx = -kc +s * ]. }) T8 m" @8 X, r
* c5 q8 G) _0 ]
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 V8 A/ v! `. p3 m! H
do you mean it is:
9 n4 W, Z% ^+ o7 Z: |( I! d$ ~& B2 h" L: A2 l# j) h9 q
d { (a+bx) C(x)}/dx = -k C(x) + s
3 N" H* |! Y* C) E. `7 H8 j) L) F6 O% M4 m* r
where a, b, s are constants, you want to know what is C(x) ?# \9 s: \* _; r: L4 U; i
1 k# m" I4 y* E2 {
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:/ |& G; Q7 S7 l! L8 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:- p  [# O5 x, i! k% t' s
9楼的答案有点问题吧。5 D* n8 f, k! I
您说：
1 F1 g% i* }# m: q4 f7 [  G+ H+ S  g- b2 Z
d{(a+bx)*C(x)}/dx =-k C(x) + s
9 O2 M- z% A: E) N+ |6 L; xso:
0 i) O) `7 Y, G% bbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s( l) h* x( X& ?' x6 T) S! ^5 e& _7 P5 C

' c; [4 E- i! q* e& R也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
3 V9 c8 c$ y& R9 ~1 B% n但这是不正确的
. f( Z$ h: x1 A- C6 kd{(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- T/ }7 ?2 R! c
...." k+ r& F6 g" O2 J
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:& T3 _, J" K' a! e5 D9 z3 [! g
这样算混淆了微分和导数的概念。
$ J2 { t8 r: R- O( q) d5 G* M$ w, Y: O$ G2 V9 Z
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. N) N# p. C6 Q& V4 u
/ j/ W1 d( P' j
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:
; E) S4 J+ `" q: F这样算混淆了微分和导数的概念。7 m0 \+ h2 A; V& B0 g: M

- G+ Z0 |7 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
5 b, b2 H8 _/ U7 R1 v" a" s- `
n' X6 n) o$ [! xAh, you made a mistake: it should be:' x* _7 k# K7 f
" u$ J4 e5 @6 J2 V. R0 K. e# j
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:/ Z v5 K0 q" f' r8 T. ~
Yes, there should not be " ' " after "f" and "g", it is a typo. :D$ {7 W# f- n) g }, R
; @- q3 t, T1 v+ F* t
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:2 S& @. d1 H0 [! `& I
* e( @* J: K2 p0 `% @9 h
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。
, n: s7 Y" A! i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:+ [7 m% _: @% d$ H6 c: W& U, d: 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？

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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