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:
0 |2 _' S% H# L& d/ x请教大家一道微分题，多谢！
3 U9 H8 D; k, Q' L6 G# V; g/ N
9 B, v* r9 K; @8 q sd [(a+bx) ] /dx = -kc +s
2 ]! V4 F, T& A. m( _8 \where: only x and c are unknown, others are all known, requiire c = ?
4 h) B: p7 R3 e) o( n; k' k! V# B/ E2 o- y6 n! b( i
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:' s6 W! C: C5 k3 C2 a9 h5 l2 G
it is too easy:9 O& r$ E& K" c- u) g: s6 G6 G

) r! G3 J: u. I$ X8 d0 r; bd(a+bx)/dx = b  K$ r) n. l8 L4 z

* P4 Q* b; L: `so
9 H; P" x7 S3 z! P! [) W' H) w  H- ]
b=-kc+s
8 c3 ]# L/ t5 k' M
3 K% G: S% [- C8 w' G9 dfinally:  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:
* d$ S5 u6 J7 L请教大家一道微分题，多谢！
# _$ T) [9 R; K% [* ^0 |/ y* h  t" V/ w, n2 R3 o
d [(a+bx) ] /dx = -kc +s : e3 C' \9 V7 \! ^' N# n" t
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?) I. {! H$ V' `* U5 c

6 }+ }+ D; S' b% o多谢了！" e  R7 F: Y7 Q  _7 B5 i
6 W( n) u8 s8 u2 F
[ 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:
* M9 l' ]0 s# a$ l8 S8 asorry, did you post another question? your statement is still not clear.. ^+ z" a. ~4 k) I! W4 J' `

M8 t1 n# i0 h* Q* a. G" pa, b, s, k are constants? c means c(x) ?
& i7 O% m* q1 Z$ h+ [- Y% D" f. m- K& v2 y, I [/ s
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:
- q6 W! V! x) y6 ?1 n对不起，写错了，忘了变量c,应该是: O! q* B! I* r& H# }
d [(a+bx) c] /dx = -kc +s + N& y. U" E2 e5 P; {7 I
4 D+ u1 z2 U, i: G
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:. k- e/ Q: x0 E  J
do you mean it is:" L6 h7 w! S# i* c2 `8 ^

* K$ t* @! ^7 q# K! D. T" Md { (a+bx) C(x)}/dx = -k C(x) + s) c8 R  R2 e$ L2 X( [% g7 k

% }  C) W- U9 J8 H% r0 K# Uwhere a, b, s are constants, you want to know what is C(x) ?# M6 J2 h. o# j) z) O1 s! `" U7 n0 L& r

; j. o; ]0 ]5 M. q, v2 eif 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:
; C, k4 T, B/ h3 \! c$ u3 g) u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 19:34:5 q8 d  V  q2 Z+ w' ~
9楼的答案有点问题吧。
0 \% }& [6 W- P2 D+ X" o您说：3 g# d) H4 {( ~/ j4 L
* u3 b4 h, U) b6 \
d{(a+bx)*C(x)}/dx =-k C(x) + s
9 [* a; S5 `0 O: C6 j" k& Q1 {* gso:
! N; K: l" E9 ?$ VbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s! q7 T+ r+ s# a- A

( }) Y& H0 i, ?& W& i: l; s; F, s也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx3 `  g$ a& u! D7 ?& v  p
但这是不正确的6 k$ e, ~4 \1 ~
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8 j$ F( U$ E N* @
...." x+ Q4 f! _2 ^9 m$ {& e
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:, k8 @, o% O' `6 W1 w0 `" {
这样算混淆了微分和导数的概念。
4 k3 y# K9 I4 k) F) @
: M! b. _. o8 M, j- m) Ld(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' i) N' `( Y* w2 H. D5 ~

# G- Y* L/ D6 n' pIf 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:! ~- b& e) q( Q: v7 ~9 Q
这样算混淆了微分和导数的概念。
% F% f2 e& @( _) P0 S# v
3 ]% P {2 ^! M! T6 \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7 p4 b+ N7 ^0 u. |

/ P" J4 Y/ K; v0 @' ?' |Ah, you made a mistake: it should be:
- u ]+ U& K: ~# y. u* c0 p0 F, d
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:( ^7 u; J2 B* `9 Z
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
- X6 q4 Q3 F0 d/ S5 Y! F
: _8 O/ H4 p2 Y" B% zBut 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 Y7 _3 r* n X) r, b z: s+ c. Z
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。
$ l6 @* K9 R0 o/ D1 p- 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 23:24:7 [+ o3 l9 d% I
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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