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 n, Z, H* |
请教大家一道微分题，多谢！
6 j( F: [ h$ s D+ {- I# z9 ? ?. E2 p0 X: D
d [(a+bx) ] /dx = -kc +s
5 p' }, U1 m8 U/ g9 Zwhere: only x and c are unknown, others are all known, requiire c = ?; k r# l V. C# G6 \5 Q
7 W* D# G5 i. y4 {9 m
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:
5 F% v# H! S @( \it is too easy:
& P+ G5 o3 {" V/ m2 o, B4 V9 L$ [' I6 o; e+ S8 }
d(a+bx)/dx = b
2 K* `3 Q( B% ~* N! f- R( M: w: k/ T# r" \8 Z
so) u3 V" [) d1 ]# V( m3 R/ V" c
: B' N. L# c/ G5 u
b=-kc+s" v/ g( i7 C% l6 q
0 Z: `. e" |. L9 O4 A3 \0 y
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:& J' b9 ~/ y# t0 f/ }; x. y
请教大家一道微分题，多谢！
1 M) z7 i! m2 u* t+ Y) ?
* F# b, y0 h. D0 ]# L8 T+ Cd [(a+bx) ] /dx = -kc +s 7 P6 q7 N; S/ a; E8 |# a) g# R
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?2 `$ {# i. ^9 i: y7 ~
+ k+ e7 L+ W! a+ I
多谢了！
3 N& \* b' z& F, F/ |
( B0 r) L1 t6 e- u! y[ 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:( o8 P! [  h' ^: T& v! e
sorry, did you post another question? your statement is still not clear.
; X- }, I3 Q8 Z8 m" |  Z* m
/ s( h3 x( D3 J( q9 D% |a, b, s, k are  constants? c means c(x) ?
8 L7 o6 }4 L# p& X" X9 l
/ {9 u8 m% \, q8 V6 u& {/ @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:* O) ^4 s, d7 E2 C( Y& {
对不起，写错了，忘了变量c,应该是
7 |1 d! {' A2 ]! G) `d [(a+bx) c] /dx = -kc +s
1 J' ]# v- a0 p5 g }" X3 v0 z* [, b5 B4 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:6 c. i; C$ u" F, }# m, A/ I
do you mean it is:. g% z! E. U" l& F
3 u1 l" v4 D* V6 D! M# ^: o$ h
d { (a+bx) C(x)}/dx = -k C(x) + s; g2 c) q7 T1 I, j
0 m) J$ v4 ^& j/ J+ k
where a, b, s are constants, you want to know what is C(x) ? N7 ` Y" g" e3 J D
7 S+ H. h3 R7 F: Q6 O* U( C
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:. V( n; a$ J' L$ T/ D! B8 T* C
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 19:34:6 B2 E1 [  s; E
9楼的答案有点问题吧。2 m  B/ W- z% w+ E4 p/ b4 I  f  H, C
您说：8 d- ]8 L  _) N

6 |+ K! L; Q) W4 F. hd{(a+bx)*C(x)}/dx =-k C(x) + s- E2 J+ N5 A1 T  ?
so:
( l7 v  v7 W/ Z7 X/ B) J3 wbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s( `1 t! ~* b9 w# W3 z

' [$ g. s; p/ _  i也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
- d" l6 ]/ T- H8 Y3 H" n* ~8 H但这是不正确的
1 e( D  Q/ ?# v0 d9 V8 td{(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6 H! A' n. g+ p) D# T; g$ X* S! @
....: ~* D. E* U! ?
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 20:30:9 Y- S. \ M: V2 U
这样算混淆了微分和导数的概念。- W P9 Q1 T( K F

) @8 ^3 i$ @5 O7 Id(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; ~! C/ I8 U5 e
$ T: P8 b$ _1 W( Y0 G+ t
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:
& Q3 I/ j) R/ Z: E1 G& I. H7 l ~这样算混淆了微分和导数的概念。" Y& s; G, S5 Y" T* Q- @2 g( _

9 f3 c% I+ b2 a1 v- {# d7 Pd(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
+ C& K0 X; I9 W( L$ N# u
" i1 `! k% ?: g9 g% n" A2 AAh, you made a mistake: it should be:
6 S y! p) H2 G( m3 [: b% i- G, l! S o6 `
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 t5 H4 C+ M; x8 \$ uYes, there should not be " ' " after "f" and "g", it is a typo. :D, ~9 }# j( Y6 l

4 h& ] @" K. BBut 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:4 `# N7 O* t+ V5 y* c

/ A8 u$ Y+ p ~5 R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。$ T2 Z5 i: v0 F- f/ }
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 23:24:1 f* L8 z5 m& H; x: R0 q: 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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