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 z7 D4 |3 w- e, }
请教大家一道微分题，多谢！
( ?; [4 [( ~3 T; T* x# K: e+ Z- e6 E
d [(a+bx) ] /dx = -kc +s 8 X' ^# ~$ r$ w% ]3 ]
where: only x and c are unknown, others are all known, requiire c = ?
2 Q6 G2 u$ J0 Y# W% ?( ?
% w H9 `# d- 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:
+ H9 ^3 Q; }  g9 p; Dit is too easy:
0 ~  J& O) W; q, ^% ]$ I
4 H1 [/ N; s4 c( ?/ y& d. ~8 bd(a+bx)/dx = b
/ l/ w2 `) e/ s, @; X* x4 k; |0 k* S. Y: g+ P
so: x& @5 y# H. e" |( Q- A" ~
( P0 k8 I0 k+ r/ f- D) o! F8 o
b=-kc+s
) W, z( i* W$ q" X% |# C; w9 w
# Y  G4 }3 e; S$ a+ ?8 @! B' @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:& M( |" a- O2 |, H+ }, G$ f
请教大家一道微分题，多谢！
8 {' I9 C: W* @1 t/ \" [2 |9 o; D- g% R' ]
d [(a+bx) ] /dx = -kc +s
; w% R* d6 ~: r! pwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
. D( t7 y. F) ~5 D0 v- t- q( k6 `( C
多谢了！
$ B( D% y/ q) L1 r8 L4 x5 n
5 ^7 Z+ t, u! f, H- R[ 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:# ~+ K6 n. \- ^0 W& H
sorry, did you post another question? your statement is still not clear." a7 k0 f+ T6 @" C

. V8 r; l2 q$ [/ [: _6 ?0 D/ Ma, b, s, k are constants? c means c(x) ?7 x9 m( _9 ~! i0 T1 C
3 [" ~: ~% e( ~0 O1 ^& f
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:
6 P! U& x" k9 d O对不起，写错了，忘了变量c,应该是! ^# G' q$ j% A
d [(a+bx) c] /dx = -kc +s ; n ~4 V% f( ?. S& J s) X0 g
7 _! P* ~; Z6 }6 }; q6 ^" H
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:1 R8 A$ [! L9 ^ W
do you mean it is:
+ I' b0 n3 p, w0 a" I7 L+ o. |9 H3 H( ~; x0 k7 T* `' g
d { (a+bx) C(x)}/dx = -k C(x) + s
8 W3 h6 n9 Y9 W3 ^0 @7 V# c ?3 s' V% A% ]% p! W% T0 Z+ x
where a, b, s are constants, you want to know what is C(x) ?6 d0 r( a/ Z0 x4 {& `

$ e, q" b& k2 Qif 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:
- Z w" c H' V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:
2 }! Y A9 w8 a# }9楼的答案有点问题吧。& z2 k: s% u/ W! S2 b ^" Y3 x- x! }
您说：
8 Y8 ~8 Z. r4 V/ d( c/ ~& R; u8 t& F6 Z( a9 S% A
d{(a+bx)*C(x)}/dx =-k C(x) + s
/ [( }( M, J0 t! j' d8 X( jso:
& x. a5 H3 l: }; t* F" kbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
) D; c$ B2 q. w4 R- ^) @5 N& D! B0 L9 Z- \+ J0 U9 x, }
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx) ]; k/ K2 m: k$ ]
但这是不正确的7 N$ h0 @+ C9 c# }( ?) i
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; ]* O: n. @" _+ b" S# J$ }
....
" Q. R; |% n! f) Nd{(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:! g1 f9 c- Q8 e9 o) i& q# p
这样算混淆了微分和导数的概念。
& d9 I& Q1 X0 ]$ [% S# B( F; ~. o0 R# d& q
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 P: C% N5 z8 k- h$ m6 [2 n7 C6 R1 H; | H' Y0 W# ^
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:
2 j; q5 f; Y4 Z. J5 V这样算混淆了微分和导数的概念。
A: Q: `) A- u9 d0 E1 u6 a" g) c7 [( v
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2 L8 R" g( m# ]( w% x1 o
) z5 y' n+ R! w8 i/ v$ d; ?# x
Ah, you made a mistake: it should be:, D) }! t9 o8 H+ s# @
. v4 f. ~6 \3 j' i
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:
& T5 M* Q% U( _7 nYes, there should not be " ' " after "f" and "g", it is a typo. :D
- G6 a! q% p% D
4 S) U+ D3 X" k7 jBut 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:) I7 e9 w! O! h9 i1 B
3 N9 \& b; n7 C/ F
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。
7 C! G( s$ o( o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:
; r- w# ]! C* x- W4 e" W+ 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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