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 20:41:0 i1 f$ g& y% p% |4 h
请教大家一道微分题，多谢！- K2 g8 `% }' G; k; F9 O

- `" k2 M  u* U% X# qd [(a+bx) ] /dx = -kc +s
0 U3 I3 q) n  h" `- Rwhere: only x and c are unknown, others are all known,  requiire c = ?: a4 P1 q7 r4 Z; y" g4 D0 K3 ~

  k9 {4 R( ?  L3 H( k# E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:
& p# K0 @" \9 {" x+ W8 }it is too easy:, F7 c. q& A' }& D6 K1 p9 K

! }( T8 p% F! j  A8 Kd(a+bx)/dx = b
% ?: f  N7 p" f& N! ~4 b) Y/ r) a' e: ^3 r# `% t0 z
so
. M3 T5 |5 r) U  k1 l: W/ e  Y6 P
" ^0 b- O3 I; O1 \6 g- Gb=-kc+s- q: M: d* Q& o6 V/ r4 u
  N' C7 E, q" v7 V  a% J& N
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:* Y4 w) L% o8 v# f7 u
请教大家一道微分题，多谢！. }+ k0 y- N6 U( p8 U' P6 w

+ K9 v) e' }5 C/ }: l: _0 xd [(a+bx) ] /dx = -kc +s
9 n& x( R5 b# q; L; _where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
2 {1 \/ p- R p
! s, j! [3 Y+ t6 ~0 Q; w; O多谢了！# V2 N/ v6 a$ v

* \1 U0 K" [+ i0 [' z[ 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:( }+ Z+ A7 C0 W' [
sorry, did you post another question? your statement is still not clear.
: j" U8 R& W+ ]3 n5 r5 O! H" H
1 T5 {$ B4 H+ ba, b, s, k are constants? c means c(x) ?
F9 [8 Y- F5 F& U. B, w
2 G& W2 g; e0 n0 ~# {# A- S, Gplease 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:
2 L4 ~& e- u+ g对不起，写错了，忘了变量c,应该是
' B' P( Y& R! {% X6 F+ rd [(a+bx) c] /dx = -kc +s 6 K' _+ z' A- [2 l6 f

; _+ |- [- t6 Y, V9 m# 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:: f4 `7 M: S" t" I: u3 J4 z
do you mean it is:/ \5 S8 C8 _0 s" U; l
1 e; e9 G! F* K2 [6 a
d { (a+bx) C(x)}/dx = -k C(x) + s
+ L- x- c0 T9 ^$ \3 n; ]) B9 v6 x4 f5 S3 X+ k
where a, b, s are constants, you want to know what is C(x) ?) R; z6 e6 G# W5 N' A8 O* Z. p1 s) J& P
& O/ q' {; U5 n$ [ K
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:
3 S+ L* ]. |' x* g0 q! V9 S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 19:34:7 d( v. {+ }% Y) V' M4 e/ @4 M3 P
9楼的答案有点问题吧。( K3 |4 L; m8 `! _9 `' x
您说：
0 }, f! A8 h( F7 o" I! A9 e, u$ N9 y3 `% S4 x9 c. A7 f6 c7 n* E
d{(a+bx)*C(x)}/dx =-k C(x) + s
" C6 Y/ Q6 h& p" n9 W) w0 iso:: R" x6 c1 v4 W: R+ X: m8 s: p
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
0 P! C- F2 y& X8 d1 I7 ?
; d* \# d# X9 l0 V也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
, f7 b- L; |+ z# i! S2 q$ m但这是不正确的9 V w6 e. T8 E/ F" p
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
9 F0 L9 m' g- O2 u" B7 S4 s5 U7 |3 v....8 `* [! v" D- Y& A" q* e
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:. J3 M5 Z: B7 s# d; {7 b
这样算混淆了微分和导数的概念。. ~+ Y7 e2 m: ?; x
6 A9 F6 h6 \# c8 W% d; S2 i
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
9 o' a- b* l) n- |& j" g1 n7 P0 b1 |
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:
) l# J3 f0 n) U" F; Q6 W) K这样算混淆了微分和导数的概念。, J* b: q5 m$ |% i! ^

- J! F: X' R+ {, q3 h/ i0 [5 Wd(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& A; Z% Z* ?8 l# X
. U8 o$ W8 \' Q5 G) ]! t
Ah, you made a mistake: it should be:
) _' H( R) a& H3 R$ C
0 U7 y4 U2 F$ A9 f' I% @! ]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 20:51:7 s+ Q. M7 d% s6 x
Yes, there should not be " ' " after "f" and "g", it is a typo. :D8 T8 R# J) |4 J6 [, _, ^0 _4 U/ o& z+ _- g
% j" V9 J- Z) ?! e
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:) ^1 H/ W& D3 r/ p! t
# ^9 h w2 x" V% J: 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。
4 `: i( W4 g/ \/ x; ^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:
0 M3 J1 L0 {1 _" t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