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:" m5 x3 F, r# Y, W6 l3 }, O
请教大家一道微分题，多谢！
. J/ _' E @+ {1 M: J
3 f! m0 O8 s' g! o( Md [(a+bx) ] /dx = -kc +s 0 |% y+ y# W9 k: I% ~
where: only x and c are unknown, others are all known, requiire c = ?
. C' b# U2 W) m! O) D; G
- O' \/ e& l" z% z1 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:/ S2 F& Q) Y% `  X0 G
it is too easy:
% e1 \7 o0 R) {: b+ Z; E' F0 K9 e/ }9 ^2 U9 A0 B0 Q1 k
d(a+bx)/dx = b2 _8 O0 z# a- ]' N) m( C
- N: X$ I7 ?, [* j6 |+ s
so) U$ _5 N! h* k6 E
( W; K9 z# Q) k2 }  X6 L
b=-kc+s) S$ M- l+ E) y4 o
$ w4 l! R# V) r" L7 B* l
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 20:41:9 U1 M7 U9 i7 w  ^4 `
请教大家一道微分题，多谢！/ m+ K: B0 [  ?
* ^/ d' @1 \! u( U( j0 m
d [(a+bx) ] /dx = -kc +s ! k' k1 C8 O, |/ D$ b, ?3 X
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?; n' }9 E& S  Y8 m8 K* P2 S

0 Y: D6 ^. s% W; r0 d3 d% c0 k多谢了！3 y9 }! \9 {6 n8 T8 F: u
( C6 U) ]* T2 e( D8 S% F5 ^. p9 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:
3 j: ?" @8 W3 i9 Zsorry, did you post another question? your statement is still not clear.8 Z' X1 f8 P& p+ w3 U0 a& z

2 l- k& Y; Z- z- N9 xa, b, s, k are constants? c means c(x) ? @: t8 s& \& ^1 N0 K& N& q/ J

% _( w3 X0 d' K! @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:& Y9 p1 |& g+ [9 S5 Y* G
对不起，写错了，忘了变量c,应该是
$ J) s" E2 D6 g/ Jd [(a+bx) c] /dx = -kc +s 3 Q7 x9 W* m* F/ H6 f
4 G& ~* r2 D& e& b
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:
# ^( Q: O  A/ A/ J2 K, Ddo you mean it is:2 F: N9 [0 @1 X% h' ?
5 H/ r: `! ~2 O4 J# D0 t
d { (a+bx) C(x)}/dx = -k C(x) + s
8 Y" g  E4 K# C
6 ]$ L: s4 f# s/ g: Z2 H' ^4 P' k) f5 ewhere a, b, s are constants, you want to know what is C(x) ?
' O3 ~0 U! N( h7 ~  ^
) g& z. p8 V- @" }; Cif 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 `5 M$ v+ V7 N+ @" B) u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- U% u4 }; _- `' ]9楼的答案有点问题吧。2 w6 [7 J. l5 L" k! Z( M
您说：
2 A" L2 M+ ^: E+ ~; i* d5 R: Y3 d9 ^$ w9 \( t- K
d{(a+bx)*C(x)}/dx =-k C(x) + s. P* c3 b! |' h6 ^7 r$ @
so:. w' O6 I* |1 F1 ]2 {
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
G& J- [. V7 c! _8 r! R
$ e& k2 L( d, a! |; T. X2 b3 p也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx4 |6 D8 s+ T: |. _8 A
但这是不正确的
0 f$ J5 J1 K# \0 f. }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
- M2 _) U% Q, N2 r9 j....
0 K7 n; t- D9 b( O6 Z% e) x5 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:
6 H5 y% P: |# U$ Z这样算混淆了微分和导数的概念。 O4 I* H, f- x+ P- w7 M9 E+ @

9 A& [( {' n U9 @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& f3 q( A. w$ @( K
; ~: P9 K1 ]3 h; r# |
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:) ?/ Y) C& x: T4 x% u7 D
这样算混淆了微分和导数的概念。% `! N" }+ R7 ?. p" ]

9 m$ z- I4 n! 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1 x4 A: j9 n9 t" c9 E& P% G' V1 O* C
9 k% {8 N7 N4 R* X* s
Ah, you made a mistake: it should be:
; s4 h3 Z1 R1 \% q0 d5 [5 Y2 Y* {+ q2 X+ h- T4 C$ v
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 20:51:0 x# a! [6 n, Y; ~( w
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
! g# }6 {8 r& Q% h
4 C3 C3 w3 p& A. y, \* c( c C( TBut 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:
( } B% ^1 v4 e5 s8 g: u7 B+ g3 h5 J) b# E3 @
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。
1 z0 u, ]& F3 v8 D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:; K+ b! u3 M+ K8 y: r6 \
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