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 X" R$ p' O, U5 u/ C
请教大家一道微分题，多谢！
, k; p9 n+ X" o
8 [- m, v3 d; e! A5 yd [(a+bx) ] /dx = -kc +s 9 N" ^3 z2 ]! H- i1 G- S9 b
where: only x and c are unknown, others are all known, requiire c = ?
) g. n- J* g5 g9 A5 p% o7 V8 D& u; c) W, N' g
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:8 b) l3 y% Z; N
it is too easy:, ^5 w" |" i" R& h# T
% x1 K. r$ w/ A8 Y, W4 P4 Z
d(a+bx)/dx = b6 ]- Y4 u0 H9 ]& L

0 k5 q0 i' _% f1 Q1 G/ G( Iso: a9 ]& c4 u9 h/ |8 \! P+ w8 `

! ~$ g6 Z o& i7 ~4 b z9 Gb=-kc+s
# M7 f- b2 b, i
) K) L' z' J* [( Sfinally: 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:
% E6 i% d  _: {" B: ]  |请教大家一道微分题，多谢！
; _- \2 u  h  _
. _6 z- Q8 c+ id [(a+bx) ] /dx = -kc +s
8 e# ?/ {( P2 F( twhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
% J4 l/ j& t; G6 C& `$ q2 O9 n, A  h  X9 m( m0 C
多谢了！6 |2 r6 s2 q$ s* S  C# V

: r( H8 U; c' X6 Q6 L# O5 N2 ~. u[ 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:
" T2 U# e% q4 ~5 i( s8 p* Zsorry, did you post another question? your statement is still not clear.
6 J9 I9 i( v2 a$ Y) w- K8 l2 \+ A8 m
6 ^) X/ R% }8 U- c1 Va, b, s, k are constants? c means c(x) ? Y% @7 ~! z5 l1 B& J
3 e5 a2 e2 h0 t3 ~) h6 a
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:
# _4 U* @2 s: I# u' P% D对不起，写错了，忘了变量c,应该是$ M L: } ~! M% p. ?! m: Q6 P
d [(a+bx) c] /dx = -kc +s % Q2 {" h! c# l2 i

' l ]- l* V' e6 j5 K Q$ H. [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:' |- ^$ N# ^0 @( V' O+ s/ O, c. L
do you mean it is:' |; k* N; l6 b0 t( T$ n- D

) g, O3 a. Y" P# w3 _4 K8 J- Sd { (a+bx) C(x)}/dx = -k C(x) + s# r* Q& Z2 A! S5 u, o( _% W

' q2 R! G( U3 r1 p" s* C! xwhere a, b, s are constants, you want to know what is C(x) ?
, s7 z4 _, P8 V8 d2 ?& K3 v ^$ H; ]* \( w* C E; k k( c
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:% I- E9 [( s* U+ F; h2 P8 r0 T2 l
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:% I7 `6 u+ i) W
9楼的答案有点问题吧。
+ z2 @5 q& v/ {# a9 [, q您说：
* \* ?5 g/ O! w0 w' `+ m/ k. i# X/ \
d{(a+bx)*C(x)}/dx =-k C(x) + s5 _) M: Z/ u& w6 u3 k
so:7 j; G0 S# o) m
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
% G' A5 ?( @9 b
! _) O. Q; a% [- J也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
8 E! g. j( N/ A  c但这是不正确的
  H% r# E  `7 z1 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
q/ z( c9 v: \....+ A4 a& Y+ w: p* [5 V, b
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:
- [. G _5 m7 {5 K' T这样算混淆了微分和导数的概念。
" [' {) u4 M" p0 A/ b" @
/ S3 j4 |' l$ G- ?+ C+ v2 N5 q1 \* p4 rd(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
7 {: C6 q) e: E8 k! D- ~ K' W" N; X1 M3 @
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:
4 J# f" l: h' r' B3 v' s: _; Y这样算混淆了微分和导数的概念。
) i7 ]# z% M$ L. {& {8 _6 S& P4 O* a0 [$ b. y7 ]# }
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
" h# j1 U) `( g4 a
" o4 `+ F7 y7 o) rAh, you made a mistake: it should be:
8 X" K: A9 Y! x* e" O( r" ~! t' @
7 B, m$ _1 C& ?1 |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:( X) I1 y0 L3 ?* o4 A3 v
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
. e7 Z1 K4 `9 J9 L: x
% n% J) x y1 L2 u/ t6 pBut 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:) _5 \9 f$ R5 W H2 C- N1 \
. ?) l1 d: d3 r) ?: s+ e8 I! @
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。, ~' p& C9 s6 i- V$ _
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:- A% b6 x) V# v$ [: D
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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