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:
) `8 ?4 s$ F4 D% H请教大家一道微分题，多谢！! P; T% j* J5 c: {

/ q# l" V! G2 w7 Sd [(a+bx) ] /dx = -kc +s
. l0 o3 P. Q8 K8 O" `% k5 m! vwhere: only x and c are unknown, others are all known, requiire c = ?
: E6 S! t3 S( }4 X
+ r- }6 ?7 R0 E; y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; w8 v1 m4 U3 _7 X4 f3 a% \  ^it is too easy:2 b: w2 V; |5 z/ j

& n4 g: o+ }6 O7 i4 wd(a+bx)/dx = b1 r& Y( }2 d3 n( \' e  }

4 o; C" l2 O& s: sso
6 b  [! a# ?0 Z8 h
. g4 Q7 x2 d/ @  [6 k- hb=-kc+s6 m. x* X) d, x- }/ q" u

  t/ @8 D- k* H4 a/ _8 Lfinally:  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:
; [( D4 \/ R  j* v1 `# h0 n# F请教大家一道微分题，多谢！
9 ^2 v# w5 J! H0 A4 R2 U' t$ k& _/ K, j% v/ C2 C8 L
d [(a+bx) ] /dx = -kc +s
' T) n/ [* ~! {/ n% Awhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
3 F$ C8 U5 ?" n1 Q4 L
3 I. Q% o5 N9 s/ T3 J1 x" l5 x# a多谢了！
5 u4 L7 a; T' t+ h- _  O. W' r
) j3 t* N2 j! w# j0 G; i[ 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:6 S; K% I( t& |
sorry, did you post another question? your statement is still not clear.3 E3 A# L* h0 M$ M% Y
6 ^& K2 M' J; R6 U+ ^6 E( n3 h: `
a, b, s, k are constants? c means c(x) ?
& z& G" ^- e0 C7 { y" q1 c+ Z6 J0 O- O3 ?
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:. u4 _/ v6 M; T8 d: i
对不起，写错了，忘了变量c,应该是
! X2 Q, w* ^8 K# ^d [(a+bx) c] /dx = -kc +s % q# W* H' W' y) g

, {& m5 O: e* n$ @; a t1 X$ X4 i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:, J" U/ K+ C8 r; s" j% e' X# X0 ?
do you mean it is:
- X8 p- p* L5 m# B$ u4 Y
- y- }- r* }: ^* p/ od { (a+bx) C(x)}/dx = -k C(x) + s
2 R8 @0 W+ Q0 \ J+ x+ ~9 [& ^ E6 E+ T4 F% N' q% x# C
where a, b, s are constants, you want to know what is C(x) ?
! d8 f. ~' {$ ~. q9 e+ l( a& v& w0 f: Z1 @ S1 D8 Y
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:( T7 q" P6 j) N# 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:: O5 n$ F" P# K
9楼的答案有点问题吧。
& I3 w' g( o) |; `( X! T0 h3 u您说：
% k' C1 u) r, T& i! V8 m
, w- x1 n3 Q1 L+ M9 }d{(a+bx)*C(x)}/dx =-k C(x) + s
8 |* e( C4 [9 ]0 |so:8 o5 U$ b2 z+ D9 |
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s: k# E2 n( C; d. V$ H% G7 D! H5 M: x

n! z4 L7 r. Q. j+ A9 j也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx' G6 D, U5 \3 z
但这是不正确的
: M/ s1 B2 `, @: s1 p' f4 y+ id{(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
5 L2 g# ?; F; ~5 F....
1 `+ `7 C2 a5 _$ \; o7 qd{(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:
; x2 ^- J. D6 ~, c" z9 W( d这样算混淆了微分和导数的概念。
) ~+ U6 d+ J* o2 N* x1 W& a* J$ `
8 z B1 g4 o' T& t8 l" @! ?' l7 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
; ]& b* L q! ]6 w1 |. Y* K; u, Z% \/ u; h3 r M& a
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' s8 c5 v' [* ^+ T8 X
这样算混淆了微分和导数的概念。0 f7 h2 |! s& i
8 b1 b" [' y* q. E5 m' 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
0 B" z" {! [& o- S7 }, I1 G& Z. s' F5 q' Q$ k! z. g
Ah, you made a mistake: it should be:
" l( @4 C2 q: G' v! b K5 ?8 ]: d& I7 y0 C: S0 L
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:
+ C4 o8 c# Y$ N% P& D3 g9 `. u% X' QYes, there should not be " ' " after "f" and "g", it is a typo. :D! @5 j) R4 X4 R$ M. Q9 i. E) ]
! V4 C$ f/ A* W V- `0 N, a! g# X
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:" j5 a3 g" R, _) f

. V% [* l$ U# O# W% D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。
" m( e9 `2 \8 R% s* p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:
& h$ m" Y' G8 k& Q' Y) \* P' y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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