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:
6 G2 V* t: u6 F4 o: O请教大家一道微分题，多谢！
- E' [5 L4 Y- o3 `! |/ E' R$ G
d [(a+bx) ] /dx = -kc +s
- ?/ Y+ l G8 A" ]/ y3 ~& a m/ Rwhere: only x and c are unknown, others are all known, requiire c = ?, l9 l- U! U# P8 P( ^
4 O6 K+ `. ^# A% l3 [& Y5 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:
# m1 g9 y3 N$ Fit is too easy:) q. v# P7 J7 [

9 c# N: [+ e8 C3 X* b- wd(a+bx)/dx = b# ?; i0 O& I/ f. A( s: \& k  c

' f& V( Z8 k& u2 ?% k$ c4 D3 hso
( q, C) J! D& `8 t9 H$ L, t, d! m7 Q6 x5 P% e2 \% |. V; R% x
b=-kc+s- {: \  K5 s6 G- {: D+ N
& z/ l" {6 \( d, O/ k
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 20:41:4 \% d& R9 \) `5 a5 Y; k- c8 E$ u! h- X
请教大家一道微分题，多谢！
$ K1 b1 g( K& D- n- b! e- D# I9 r) H, T$ W% Z* u! a5 I7 |" P
d [(a+bx) ] /dx = -kc +s 0 y3 r& A. M; E% G! M$ G" Z7 B
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?5 O2 R, j6 Z6 M; [) @( D
) T7 s$ q2 W8 b" |( \4 I
多谢了！
+ G+ k( I3 B5 j7 I& c% n
* W6 f) Q) z! x! N B[ 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:
0 E4 r1 @( |4 T0 q" `* wsorry, did you post another question? your statement is still not clear.
$ K6 J" n/ u8 J% R/ K7 w3 X* S% q; Y" Z- i
a, b, s, k are constants? c means c(x) ?
9 Y: P. R. |3 K5 n" H8 ]1 E1 M4 f! V1 Z/ S6 h
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 12:30:8 G9 z! \7 e. b, h9 p
对不起，写错了，忘了变量c,应该是
' Z+ ^2 |! S n4 A0 T0 ^d [(a+bx) c] /dx = -kc +s " \1 |- k" ~/ G6 e/ m* D+ e

6 ~' O5 K& C( w* [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:4 S0 `1 i; a8 s2 d  M
do you mean it is:
" D) ^, `" S" U" A  ?5 z( j
+ F# O, m) u+ I3 A2 Dd { (a+bx) C(x)}/dx = -k C(x) + s
& ?. U) |" R: b- w2 H! L
4 t' N' F3 l7 N3 G+ D( ?4 Vwhere a, b, s are constants, you want to know what is C(x) ?( o0 M  h" e- m, q, o

% s9 I: |+ [8 d6 H" m! ~( rif 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 14:33:8 w* u2 L, s# h$ p0 R
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 19:34:9 Q$ c- `: y" s) o
9楼的答案有点问题吧。
! B1 b- j/ |" Z# ]0 k( ^9 p6 H5 d您说：
- j) T' _8 T& o
% y' h) [0 ?3 n: J$ m9 w$ q, ]d{(a+bx)*C(x)}/dx =-k C(x) + s
& A0 [% B( y% D' [0 ?so:
* g- L5 ~0 |- K8 gbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
! Z" H9 D6 C: J( O- Y
2 i6 @9 e0 u: X  G! K# H8 ^" G0 Z也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx+ g+ O3 A) @; K9 L5 i% B6 E
但这是不正确的
  q2 N; \+ \9 C( G. e9 Xd{(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$ N, @+ Z2 K1 x
....
& i* K( E5 L' ^* ed{(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:
7 Y, [* k% S) Y1 L这样算混淆了微分和导数的概念。. E1 A4 D2 @! i, `

- X( N( N- c9 z2 t& |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: X' A) O2 {: H3 j% e
$ E i6 N/ @" Q2 C& D
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:
8 z* T; H2 b$ l2 S0 E这样算混淆了微分和导数的概念。
/ Q- D( M7 i4 ?# j( t7 z$ [. J3 D. |$ ~9 u- j* x! [) V$ W+ a8 J
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 R) Y: x0 ^

; e5 d8 r% F2 b) YAh, you made a mistake: it should be: r0 f, P7 q: @5 t$ I* U* j
[0 H5 T7 w% v* B+ [2 h8 Q
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:
& W4 S8 a8 s2 |; K& A. x! wYes, there should not be " ' " after "f" and "g", it is a typo. :D' n" Q9 t4 B- O3 Y2 H
% a5 \; q' J9 [. g/ Z: l& V
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:$ ]6 T/ d1 l" \; w4 Q
9 o% Z `% n( R& J5 s% z4 R6 H
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。
, c8 {. I$ P0 {5 E/ g: S, 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:
( Z) B6 w3 e& \' b ^# @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。

账号		自动登录	找回密码
密码			注册

数学高手请进，多谢！

浏览过的版块