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:
/ R: d7 S2 m. G请教大家一道微分题，多谢！+ i9 o5 z' j% t) d: E

6 l" y! l" M/ L8 G7 k- Vd [(a+bx) ] /dx = -kc +s " ]$ c4 C; o9 k9 [4 n) F
where: only x and c are unknown, others are all known, requiire c = ?' o% k7 U. i. H/ S) q

: D3 W6 T9 e3 M) O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 D( f2 t0 l& L& ^+ V
it is too easy:: [1 c3 V8 X( i. W# A
6 H* ^3 s7 x- I. P) P
d(a+bx)/dx = b
0 L( Q7 a/ K3 ^' O& @5 u# P: d
0 g6 c0 _9 H4 i$ O2 Hso
) A' C/ L. S+ \0 f: n3 S8 ~ V3 T. q
b=-kc+s
- p$ i% i. K5 a8 t; u$ w: z
6 i+ k Q; ~+ }( }! t0 t* U8 mfinally: 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:+ x: k" b& s& z9 l
请教大家一道微分题，多谢！1 O4 ^# f ]% H8 {
8 b( u4 T9 o, U$ _& M
d [(a+bx) ] /dx = -kc +s
8 c( d: W) t$ m! jwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?( V) D ]9 N( m* c- |( d& e

4 W! o: x' r3 v2 E/ [2 E3 d多谢了！
2 y6 |1 O' T: l% m' i9 \" c! G
' S( B) {3 J4 }- u[ 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:
. [( C5 a: `4 f5 E% h. s6 m, ksorry, did you post another question? your statement is still not clear.& X$ J5 ^8 u" Z( k% P
* `- v* u+ w0 v9 _# D
a, b, s, k are constants? c means c(x) ?+ X: p/ M$ h9 `, C
9 p' n4 _; R9 G
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:% t. t8 L% W5 J( d2 t$ s
对不起，写错了，忘了变量c,应该是2 {8 @( Y6 S2 |0 v/ V3 x2 C
d [(a+bx) c] /dx = -kc +s + [6 F' P& _7 |9 L1 d
% f U3 F9 a& N" ^: N% Z
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:8 G( R; M6 {3 W2 [" e* x2 Z- u* D% }
do you mean it is:3 T- }, J5 a- P5 F S" C
! W E& F: p4 }1 e/ J: |- m' u
d { (a+bx) C(x)}/dx = -k C(x) + s' `( s- Z4 @& w2 N8 W- @7 x

' Y9 r, B3 {% ?* N5 Swhere a, b, s are constants, you want to know what is C(x) ?# h0 |7 V: r# ?
9 t: `0 E7 c+ x+ }) G' 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 14:33:5 |2 g' g8 @$ U3 @: {7 U* w
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:, a/ ~. s1 e' e3 D4 d
9楼的答案有点问题吧。1 P$ t0 _4 y& t
您说：
j: M' Y: s) s( G+ O) K8 o: Q( P$ f) t4 W
d{(a+bx)*C(x)}/dx =-k C(x) + s2 s+ U5 [: o8 d$ L# Z
so:
* m5 j0 Y; j5 F" s: | x3 qbC(x) + (a+bx) dC(x)/dx = -kC(x) +s9 @. ^8 W$ i0 z) A# h3 V4 I% T9 s
4 Y: H6 r5 h6 X1 U( i
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx- d6 [# P' u0 Q: ]2 J5 _- R* O
但这是不正确的) r. e2 _8 W: _7 h- w) Q' [6 ~5 E
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- d z x0 _& O! g H% e: |3 {
....
/ B" E+ ?1 N2 s; c3 Ud{(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:
% f" N! F4 _7 z. x7 f# `1 F9 }这样算混淆了微分和导数的概念。
5 Y% S! i6 f( P5 w" g
9 d# Z; M' ^8 a; Fd(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+ k% v, z" k0 _ i8 y$ S7 c# L0 Q, k3 M9 W2 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:
这样算混淆了微分和导数的概念。
# R% z9 R( ~+ I% W1 U; t9 T2 P
% r6 l, g: Q; w/ 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. R9 [8 u' J' C, p1 o
) Y- {, J% |& R# v% o
Ah, you made a mistake: it should be:# ^7 q% Q4 R0 Y1 M3 s& K3 m

; ^' v h1 R5 m8 `: Fd(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:
5 F' h/ R7 g8 V# a, T# W/ |# dYes, there should not be " ' " after "f" and "g", it is a typo. :D& Q. X i, v9 F, R* c

0 d3 R+ u7 T9 LBut 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 @% n- N: B- G4 F& Y! ^: L$ o# }! c
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。
; V' l. m) ]7 s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:
4 j) f& a6 F: O* 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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