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:& L+ ]0 ]$ W0 j# h4 d3 K
请教大家一道微分题，多谢！
) i* t2 {9 ]/ `2 ]/ _1 _3 i5 I* \! O: t
d [(a+bx) ] /dx = -kc +s
2 z$ a0 }" Z7 \1 zwhere: only x and c are unknown, others are all known, requiire c = ?* y; z$ D j$ M% a& s

8 }+ ~4 ^6 f7 q2 ]6 B$ K+ b4 R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:
! @* u8 k, Z% u, _. O, W6 R# pit is too easy:% O" t2 B8 X( [. V5 c$ q
- t+ i8 Y' @$ U* |+ ]! d* W/ P/ V
d(a+bx)/dx = b
  B& C# U) |0 Q, s9 b. G; A
$ Y) ]2 J8 I# t0 {7 D- Eso9 I) Z0 P; |$ i6 |  H+ t$ d9 k6 d% Q
' f) n6 y4 \. T: e3 P- Q( G' X8 I
b=-kc+s( s3 R( F" F9 m: u4 r6 i' |5 z+ \

7 H* [* l( h1 x: x' Ofinally:  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 20:41:8 j  J0 |  n3 ~! U/ r1 N
请教大家一道微分题，多谢！# q# n" T  G3 _: ~
3 A5 N6 C' @6 k* k" j" U1 c8 m
d [(a+bx) ] /dx = -kc +s
. X; j% |0 f* O- g/ w5 J! swhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
" h" s* |" G, K) |/ U, u6 r3 ?! {
  {$ D+ U4 ]8 D6 ?* F: O多谢了！6 V' m) s6 P  s. R- w8 v2 x

* H) _* W4 Z1 q! E[ 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:
7 N$ j; j# _ d/ g& p4 ksorry, did you post another question? your statement is still not clear.
' A5 Z& i* I) a3 ?- A1 t
& F; S! x( Y6 r% ]) C8 Ha, b, s, k are constants? c means c(x) ?5 t( U" y9 `9 E3 Y0 i0 r6 \

' A0 Q* w. |1 x' K3 }. G( Kplease 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:
% F1 s9 S$ y/ G& J0 x D对不起，写错了，忘了变量c,应该是! v8 O1 t5 `6 d% E& |; ]
d [(a+bx) c] /dx = -kc +s - C3 G8 d% P/ n
& }: u$ ]4 X9 j r3 c: ~
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:9 E/ \2 s6 B. j0 W! R
do you mean it is:0 ?! ?7 r4 ^& W9 o1 d b3 f

' N( ^7 M9 z# \; s. wd { (a+bx) C(x)}/dx = -k C(x) + s+ P5 i- s" o0 a& R9 |8 s

5 o0 B w9 z, t" [where a, b, s are constants, you want to know what is C(x) ?/ \+ A J ~4 g; Z# P6 S9 P- W$ [3 O

% g4 n2 j7 V3 ~, Y2 o9 oif 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:
: s: e1 r5 f+ C; K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:" m$ A) V# H' j+ i3 y' t
9楼的答案有点问题吧。
+ a5 p2 w8 j2 j  S- N/ O2 [您说：
3 K8 m2 e2 j9 A4 m% e, h1 ^, O# I0 `1 {: \$ {
d{(a+bx)*C(x)}/dx =-k C(x) + s
7 T; b6 U! A6 X2 h  ~! I% A6 jso:
. C9 C/ {' v0 i6 i0 i5 jbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s, _! n" W. I  w+ g" a

9 h7 n) C# W, t" x" X也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx) g" \* b9 c6 j$ G: i0 }+ D1 }
但这是不正确的
& V: Q( w$ q. j+ |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& t Y9 n" @+ s$ }4 R3 v& a, K
....
3 ^( a( n/ t6 S' R- w# @4 pd{(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: Z7 r2 ?- L+ J4 Y% B( w: q0 Y
这样算混淆了微分和导数的概念。
: z& ^( h+ c3 G0 e' k% h5 o
' Y, X) s" f7 ?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
" S" {5 |& `; H2 F4 f# V' I- J7 E& h' R' W2 B1 _" M( c4 Z
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:
$ N! E& q# K; Q3 C这样算混淆了微分和导数的概念。
) A: _) Z( Z/ l5 t: j" [8 l- z0 y# R% b
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 x$ f( H8 i/ }$ x. k8 \) u" U+ G* t
Ah, you made a mistake: it should be:+ c4 [/ ^+ \& O$ v- V" q
3 e4 [5 n: n, W0 R, f3 P9 Z8 [
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:
% e8 O' y6 |1 S* s- b+ \5 b0 s& r' PYes, there should not be " ' " after "f" and "g", it is a typo. :D* [* v: ?8 P: a+ S% O# q' D9 c. A

' n1 R& m5 U/ `* K3 i$ ]" W% `0 {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:
( K" g3 L$ ?! k, W3 Z8 J
7 B# p( p) f% _* C9 B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。% O" m, c' Y7 L( Z
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:
Z# o$ ], {8 l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