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:
; c, }4 m- T" N4 y- |9 Y/ {请教大家一道微分题，多谢！" Y3 v$ }% z5 b- {* y! u" t3 P( v
* C# S0 m$ ~) W2 p2 L: r8 s
d [(a+bx) ] /dx = -kc +s . d) u8 t" l* A4 }
where: only x and c are unknown, others are all known, requiire c = ?- p7 o* Y4 U1 a! d

3 q' {9 d6 N! z6 A" x9 ]% I3 [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:
2 S1 l. N( n, i# R& [6 iit is too easy:: q3 M- [) t# ]. B
! u. t& L) J2 q6 \
d(a+bx)/dx = b+ O# |  T1 H2 f8 @' T

& u& N6 I) H1 }7 J' K6 Bso
6 {7 G+ y  p: s' \: x, _% @" v0 b& \- A; @
b=-kc+s
3 `0 u$ d/ w0 h1 l- O* J  K& X: j3 L$ n1 z$ P4 v
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 20:41:6 l) x% w9 l* ]2 n* H
请教大家一道微分题，多谢！6 a  `4 O7 S& b' x8 t3 ?( T
( K* I6 Y1 [8 d! [
d [(a+bx) ] /dx = -kc +s
8 l4 |: ~& D' Bwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
, N% M- A& G+ T: a4 ~0 ^" i& k8 R# _. R- d+ N6 R: l
多谢了！8 Y5 M' t  E* u" H, F+ m* Y) M. w

0 u$ ^. @3 \0 B$ I' M[ 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:
' X4 X0 [; k% i1 S* l7 tsorry, did you post another question? your statement is still not clear.+ @ K4 Z: t' ^8 T" E7 w
# Z8 a/ H( o/ V- @' N
a, b, s, k are constants? c means c(x) ?. j- ^ P" H5 w

/ |$ b+ K" }( F# s5 ^$ oplease 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 P% B; ]- R, C+ L) Y/ a对不起，写错了，忘了变量c,应该是7 X0 X" G. I# ?5 z R% k6 }
d [(a+bx) c] /dx = -kc +s 5 c) t" V" H1 E" s

. P# _* I9 [- M; q" ]$ Q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:* x4 J4 y$ q% Q# _
do you mean it is:- E8 Y: P C+ _* [* T( f; c/ R ?8 L
]5 n7 r2 x8 Q
d { (a+bx) C(x)}/dx = -k C(x) + s# d4 N# g8 b: N" q4 _" }' @, W

0 A: t, R# L9 d% K* Hwhere a, b, s are constants, you want to know what is C(x) ?
' a* q+ y/ n2 y4 d/ `5 Y% [3 C& A
. i9 m- e2 f% b: V) r+ aif 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:
) o: c( z/ ]1 J" I8 a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:' y- y% ?% q3 q* N% ~7 o4 a
9楼的答案有点问题吧。
% ?6 b1 L7 i, h* `您说：) Y+ q) T# D; C* S1 s
' y! f* A: m( t* w- ]
d{(a+bx)*C(x)}/dx =-k C(x) + s
$ N- v1 @$ b) M$ iso:
& P3 I2 I& y5 d# lbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
+ }8 J3 o( k" l* @$ k( K" q
/ X% J+ i; B7 \( G2 V也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx8 V. g' P) y6 \! |* L X$ ?
但这是不正确的
& r) K+ F* f! [3 F3 zd{(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# S3 ?0 x) Y* v& a* U
....
5 H m: K5 _# ~; n% m6 rd{(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:
2 p/ v9 D- t6 J6 b7 p这样算混淆了微分和导数的概念。
8 k, [; q" u$ F" Z
( n' H# h4 _3 h% H1 ]7 {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
4 a% l/ t: w/ g2 M/ G+ S; A: U: u
* `6 p M6 p6 `% }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:
/ I+ i# @1 \9 u这样算混淆了微分和导数的概念。7 [ q) P! p: m+ Z9 i
( P; }$ E' i% }- A( s4 r
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
- d" a. |5 J# y# u- Q
5 E' @2 X% s# D6 X. |, p* WAh, you made a mistake: it should be:
& v8 P& \& Q8 T5 z$ F
, s4 B, m# i0 Kd(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:. g% X$ |9 w6 b @
Yes, there should not be " ' " after "f" and "g", it is a typo. :D# l( u7 b1 |- V
6 \# n! e$ ^! c7 O$ n
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:
: I% e% c w2 L* Q0 `" e) s$ o7 P
1 f4 I2 E' O' d+ g1 _- q, u: [9 P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。! \- I- w4 Z$ A* D0 \. c0 w7 B
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:
) S: |; r, g' ]. }9 |; V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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