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 20:41:9 h; c5 L% p$ ^. a8 O
请教大家一道微分题，多谢！1 q6 g4 r) g: X2 z5 c

9 u/ d' t: Q+ S; F3 ]d [(a+bx) ] /dx = -kc +s
) m3 _/ A6 l+ |1 T nwhere: only x and c are unknown, others are all known, requiire c = ?' O9 X: t% q# C6 o2 W: g; [" B
) J: [8 \7 ?( s" |9 @( q% X+ U: t
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:
; H+ a& u6 |5 d' P8 R; K' h% l1 F: uit is too easy:, m! u1 c# Z! E

) p4 q* u, _; M0 \! M2 }$ t" gd(a+bx)/dx = b
; |  P! I4 I) h7 R& g$ H6 Q9 r
% o% Q, E4 X) |$ D& d4 fso- u. `8 s5 J- C  ?! Q3 I

6 m+ F, M7 [6 F3 p0 c9 C2 L+ Pb=-kc+s% S* t- N" m9 p

; f1 R% {, w7 g9 J$ cfinally:  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 20:41:1 ?9 t* m, a# R( H4 ]4 g7 g
请教大家一道微分题，多谢！
/ ^$ W5 f6 u& _; W; V1 \& q9 ^5 `$ v; x6 U
d [(a+bx) ] /dx = -kc +s
; }4 M# u. a5 B6 S5 jwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?; X8 e. G# D& g

% V# z4 A1 X$ _2 u% H4 `0 O6 ]/ B多谢了！
8 y+ G3 q) _7 A, O& N9 Y, O5 J1 J! v- u/ [0 V* L% D
[ 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:9 K4 q8 A* r' h5 Y
sorry, did you post another question? your statement is still not clear.! W& d( k: d3 Z- M2 V
( q7 J7 u s4 R9 o
a, b, s, k are constants? c means c(x) ?
9 I, P6 s- y+ C- r, w9 x5 E
4 G/ Q& h0 V) k* ?7 J6 U, C: \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:, @* h1 _( c# J5 q
对不起，写错了，忘了变量c,应该是
+ T. z7 @2 R( u9 Y. g* {d [(a+bx) c] /dx = -kc +s
' T4 H" J) d% Q9 P2 |6 N7 O9 O. }8 s: 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:6 t) H. ~5 \& u; i# P( i
do you mean it is:, M# p& U, ?& G( L$ r1 F0 n  H

% q3 [; d7 \3 \2 M! ~: rd { (a+bx) C(x)}/dx = -k C(x) + s
( \. J  ?& a$ q. ^' D+ y% [
, R3 p/ M4 [; U. q* Awhere a, b, s are constants, you want to know what is C(x) ?
/ n7 B4 d6 X6 [4 U
/ B  y/ N2 X" L5 L$ Gif 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:
( a( W8 K# T/ ?- 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 19:34:8 }4 w) N7 `+ n' K0 Q) p5 W
9楼的答案有点问题吧。- J' K7 o9 o# C6 O& l/ d+ ?* l
您说：
" O& J) S- o" V3 V
4 x# P: _# p8 U0 J/ E/ [% Jd{(a+bx)*C(x)}/dx =-k C(x) + s
so:- K B: f! A7 Q" |
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
. w: Z3 p3 \3 y2 p3 M, A# m* ^) \& w1 I( ~3 }
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx3 q6 a. a6 f. H* w7 T& f
但这是不正确的
3 n& p' Q% p- G. G# i' c8 wd{(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 s# A: u+ d. R6 c) j4 j- ]: m....
, g- c t: c+ X. ?3 \0 [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 20:30:2 M% y- X. u% o3 c% f# I
这样算混淆了微分和导数的概念。) K( Z. s6 R) C9 B4 e% k, C! e( K |

9 M. y' F. W0 kd(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0 q0 i' N" S2 g) D: |

1 d9 N( a T6 G& J& Y! G7 h* rIf 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:: N; @7 o5 L, f( p  }
这样算混淆了微分和导数的概念。
2 R4 y, u8 w) {7 C* `' ~: Y
1 ]" E* m! u2 K' U: ?* y; Sd(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
5 }* v% J% ^& U/ u& ~' H6 Y5 \0 U4 A
Ah, you made a mistake: it should be:0 E# Q0 x5 l  W" @, ?$ J
  l8 c0 |* ^4 _( e+ {
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:
3 K. E) F% j* m2 p$ {2 WYes, there should not be " ' " after "f" and "g", it is a typo. :D
+ K3 J) k) O; z. G8 F/ W/ ]9 d0 D/ B3 N# k& `1 Y3 S
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:
* o0 d" M; q% R3 d1 v0 h
( X' G2 l4 ~6 {+ M: G1 |6 C7 d/ Y) `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。
. n7 K% E7 w* R, ^, L) X& _) m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:
% D- N9 X$ X! H" s3 k2 |' @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