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:
: Q' p) c  d$ v/ j& X请教大家一道微分题，多谢！
- D. H/ {2 B& J0 O: Q9 p  G' l! O% O8 ]% A/ n
d [(a+bx) ] /dx = -kc +s % j- ?% Z1 f) P- e4 ?
where: only x and c are unknown, others are all known,  requiire c = ?# M8 t) I! b" K6 s5 t6 X

8 L( W0 \6 v- [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:3 q/ i- j' p$ f- K) z1 d' T8 B8 j
it is too easy:
' [" B' k3 m. @% q) X! B) E* Z: x1 k4 n, p7 W8 j- K* n6 |
d(a+bx)/dx = b% v+ d" K5 B" [

. W9 D, Y/ F1 `9 y0 j: _so S+ C1 P9 e M# J
4 [4 E5 k( E7 J+ s( E+ T
b=-kc+s
/ Y' m4 Z/ v2 q, E& E& A9 A7 y. U- }- m5 p) A# }; U
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:
% ?+ E0 a7 z+ v请教大家一道微分题，多谢！4 a& m/ r, i1 o

8 T, b( ?+ h9 h1 \+ W1 I/ t5 Pd [(a+bx) ] /dx = -kc +s / v6 |" N- ^1 y4 V1 x. f
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
; x: P' L! y$ g% l; l, O1 \4 A g$ m) _9 L2 B
多谢了！+ y- E- y2 o& s9 J( ~
$ |- o8 j+ U3 {( L8 [
[ 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:: k( j4 W4 A6 F" y+ D
sorry, did you post another question? your statement is still not clear.0 T- W) Z6 H) P2 p! z1 A8 q
3 ~. x8 ~. P, D
a, b, s, k are constants? c means c(x) ?5 E# @2 y" u1 D$ T, [+ f; I" r2 c

- G- a4 s8 f4 x2 a% f5 fplease 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:) T0 X* @: d0 L7 q M
对不起，写错了，忘了变量c,应该是
5 G6 w4 H, @3 e% ^d [(a+bx) c] /dx = -kc +s
, L; `) u% r, N' k8 X5 J4 V& V! l7 a+ R
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:
0 D  p) f9 W& e7 A) tdo you mean it is:$ O  y. @$ t& x: e& Y, r, i% X& T" x; e5 W

- B( q1 K/ I6 \" d3 pd { (a+bx) C(x)}/dx = -k C(x) + s
# ~: k/ F' L- @) Y1 m0 H, Q! a0 l# |, g5 X! L  ], |
where a, b, s are constants, you want to know what is C(x) ?* P; M( Q* O) G: |9 s! N4 b" x
9 O3 p9 Z4 c, E- _  G8 M/ |5 l
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:
1 [/ y' Q# A5 ~- X% o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:
, }1 H( O% {3 U% [/ p* s9楼的答案有点问题吧。- R, S$ F3 e: @8 q5 `
您说：
4 `/ n& I5 ]& v* y1 [9 E- i" k: ?  ?7 A
d{(a+bx)*C(x)}/dx =-k C(x) + s
( F' w2 p  j, f' Nso:# d$ x, z6 j4 Y
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s( M9 ~& I: N2 i" l. s

& g( }! j' E% ^- c! r也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx$ H: S& I" ?4 X3 j2 X. J
但这是不正确的
8 o; o+ S( f% X# K6 sd{(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7 `5 `* Z3 `: s% a \+ t) H+ ?# U- b
....
- x' C7 a5 I+ Td{(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:
- N& L% s* D# J) c' [/ P这样算混淆了微分和导数的概念。. D; p* B( f: A; ~4 c

5 W# D2 [ E: N) e8 bd(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( {& G" e, z! ^4 u$ u0 d* M
/ r7 @. V/ h8 v; J) g* l6 O
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:6 ~% L7 ?: n. K. K8 S/ h
这样算混淆了微分和导数的概念。
% N: m' ?7 V \# g# B j
, o9 y, X& I2 y% {. Yd(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
- v0 r2 [* b+ t5 h9 q2 b5 m& |$ Y1 j7 X/ V( |: }
Ah, you made a mistake: it should be:2 C" q) I$ ~& O. E

) S: X# n# V# |# e; V8 I9 ud(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 20:51:1 d& t: W! H+ T$ v
Yes, there should not be " ' " after "f" and "g", it is a typo. :D5 L5 `0 T: A+ d9 x, o) C; F% I' U. q" i. l
1 P5 W- f, C8 F' m
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:0 i2 G* ~* ^* z- h: r
( K4 @5 O2 O% e& z* _* k
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。4 E: A' e _4 D" u# N2 x
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:
# |& r0 O, h L6 P& W& k" ^6 T" y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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