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:
; e4 ~" u" V3 k请教大家一道微分题，多谢！
6 Y0 m. g6 ?6 J8 j3 _& H; e! t( ? _! _$ t7 a( S
d [(a+bx) ] /dx = -kc +s
# @/ Z1 b) {7 W { o. x( Ywhere: only x and c are unknown, others are all known, requiire c = ?
- k+ \8 Q" _2 O& j7 z
9 T9 c! m8 ~2 f2 q4 ^! a( b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:
/ `5 `6 X2 T2 T& N' N  s7 {6 g9 ^/ `it is too easy:
8 }/ ^8 K+ V* ?& U# f$ _0 |& [/ {6 e9 K# x( x, Y1 W* Y) K
d(a+bx)/dx = b
# b7 K, k2 r3 x- |+ ^/ {, t  p$ f
& E4 k: X7 g! k* X+ v) N9 s" M) E8 wso
# i* H3 @% m! H- d2 c, O, s+ H) r2 p8 T. Y, v
b=-kc+s
6 M# ]8 \$ A/ {" \$ x" r  W. i2 i# M8 D7 u9 V0 W: \
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:
2 C8 q& W* ^; r8 H2 j3 ^& [请教大家一道微分题，多谢！2 @' D9 D! [# y+ w$ B

) N6 ?! [; o: {0 ~: Jd [(a+bx) ] /dx = -kc +s
+ O7 n' U2 b* Ywhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function? r4 ^ o: W2 [
5 v+ T1 J( \/ Q% {, |
多谢了！* H; L) D2 C% R+ M9 @9 I
5 [& [) r% ]- p/ x+ w2 V
[ 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 d: ^( D- `" j" R# n- |5 c
sorry, did you post another question? your statement is still not clear.
* d  L2 @5 p. s0 v& q% G% e$ ?
) b' d$ M% ]! H, u3 O5 D* M% ca, b, s, k are  constants? c means c(x) ?
6 j) O3 _( z: C; r# y
  \6 W5 V5 ]- x2 I: y" L! _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 12:30:6 O: ?( e8 y! Q1 D, |( D
对不起，写错了，忘了变量c,应该是2 H0 {7 t$ ?0 k
d [(a+bx) c] /dx = -kc +s 1 l$ u5 N$ @8 h+ _6 U# j
7 h3 \; }- [$ l, M' m* x6 s( D$ |
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: u5 g l4 t- W' Y8 m5 U2 a
do you mean it is:
, [, V7 S3 {4 N! R! P& x4 ~! s, F9 N! }
d { (a+bx) C(x)}/dx = -k C(x) + s8 f+ Z" F, F5 W4 U

7 u) }+ j* L0 I( y. q( S6 |# Iwhere a, b, s are constants, you want to know what is C(x) ?
$ i4 J$ k" h2 s7 y6 P g% I; P' Y _+ f. r. ]9 n
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:
[% M8 K# f0 X# b. A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 19:34:6 A# n( O& n8 ~3 b8 n0 ~
9楼的答案有点问题吧。1 ^& w( o( ?. m2 ^3 Z" ~
您说：
! M8 F8 t- v9 ]" {/ A# K6 z, i: x7 P- G& z
d{(a+bx)*C(x)}/dx =-k C(x) + s7 X% @' B9 ]3 N9 C
so:  T, [) r8 `/ S) F9 ^( W4 G+ x
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
! \  ~/ g0 O- r! Q1 |2 ~5 o  m) ?3 H5 R1 ]
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx5 {; Q: V5 m& a$ b) l& }1 K$ X
但这是不正确的
2 S# `- [/ n6 d  @9 M4 A* z' ^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: l& E- @4 v- e! l
....
0 P) m. B- _+ z4 q' L1 |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:
$ i/ n. {+ F  I3 n* ?/ e3 Z这样算混淆了微分和导数的概念。
. e7 N% B9 O1 o5 H  C' L
- u5 K' u6 o* f8 a$ Ad(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
% \! Y$ x- }, e& P! C* r8 K! m$ e* \  Z! M
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:
2 o q0 W/ p0 T& G/ f这样算混淆了微分和导数的概念。/ j0 e/ @2 B7 N9 m% G- w
2 I& H3 s6 Z3 A. S
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/ M5 w3 O. U6 T' @$ x: {$ I

) J( P+ ~7 g% U8 ^. u" ?9 AAh, you made a mistake: it should be:
/ k! b: [; Z# t* C/ h1 d
+ T% H( M5 E7 Z: h' F8 od(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 20:51:1 v7 S$ @/ p/ E, ^
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
1 L. b/ o6 {! u% b* A: O9 \. X- o! K- Z
: W% P& B5 ^; x. d' z1 VBut 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:9 Y2 ^' U2 Z: ?: y

: [) X- u) f9 l6 U$ P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。
, ]* F. h- R: y* T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:; U/ c$ ] u" g C
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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