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:
; m- z& E' q4 r( t# F4 S% w请教大家一道微分题，多谢！% z. Q( M3 ?/ O8 i$ ]6 q6 X9 ]# k& L
( |) A Y) K- T6 [$ _
d [(a+bx) ] /dx = -kc +s ' C# W% k8 e/ A
where: only x and c are unknown, others are all known, requiire c = ?7 l9 S+ ^' R% o+ ^% p* k
$ M' b5 i$ T; I4 E. m
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:" o% J# {0 q6 t- s
it is too easy:
. I9 C" X) w/ _$ ?2 c/ e1 }5 M" A9 w& H5 G# k
d(a+bx)/dx = b- M; `/ X3 k3 U6 j
, ]9 @% D, G9 H9 X
so
( \3 p A% R4 I
% l5 r& R4 a; E8 l1 C# W2 @3 Db=-kc+s8 b( n) t7 C" B+ q) ~" ~

( C3 g" O% _- D7 [: 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:
% n, L& [6 m+ H请教大家一道微分题，多谢！
) n! ]2 @: ^9 P0 z0 E) g( J `) ?* a' n9 D4 j
d [(a+bx) ] /dx = -kc +s
6 N' n W1 D& v' R/ E& m; qwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?8 W0 A3 `; V. n4 H
. \: |6 L- l% Q( g( s) s+ e
多谢了！. ]0 P. Y+ T7 f1 F; N! Z8 _& s8 V
! m/ O, [3 `; `
[ 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:
: D' y# ]4 t; M) K2 {' m, Isorry, did you post another question? your statement is still not clear. |- |2 ]- V5 k) r

$ Q* H& {- P6 i S$ {/ F b5 `a, b, s, k are constants? c means c(x) ?! O( c8 |* k( A1 z
, k/ L( \1 y0 s9 M/ H
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:) i6 l B( _$ L4 W" C3 v2 q/ s% d
对不起，写错了，忘了变量c,应该是 r( U/ x0 _6 ?/ r& F) T
d [(a+bx) c] /dx = -kc +s
( g. q/ _6 Z5 j6 q& O: Q
: o/ o% ?& x4 {( S* F/ k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:
P# D; p0 P" n3 u$ bdo you mean it is:
- [0 {- k! w; u$ m: X" R0 n- x: q
. ~: D5 o# b- I& m9 p, a$ I$ Ld { (a+bx) C(x)}/dx = -k C(x) + s% C% S, p7 I& M$ O% W% ?
8 e) V9 \- i' L. K: m
where a, b, s are constants, you want to know what is C(x) ?
2 h* {3 `! X- s5 [% n* b2 t4 F, X# O% Q' ]% i1 U+ ^" Q
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:' v' D! m+ C6 Q0 T: s/ t! Y
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:
# U2 ^- O2 H' r# T: ?9楼的答案有点问题吧。
I4 {6 x! H' b- V$ A: g您说：& }8 Y$ E" s+ x- z5 E
7 l% L' H! ]+ D: \7 |- p! B% k2 }
d{(a+bx)*C(x)}/dx =-k C(x) + s
) H) P, _* I# i+ d: [5 W) gso:7 S; }; Q0 g* ?$ Q3 U# b5 ~. o2 A) \
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s/ K2 `( ]: S) O
- Q+ P! a' p1 W3 H1 B# X% r
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
3 p5 g# S- x5 b( M, v8 ?但这是不正确的
7 G0 z4 w' P6 {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8 k+ \% i3 x# a3 a
....
- }, o9 f5 p! _9 m9 m# [# d7 d! W5 md{(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 20:30:1 J8 h; D- N: `, t& k
这样算混淆了微分和导数的概念。! c% P1 [3 G, i4 b( [

1 R1 [: Y. ^7 ^; ~9 ?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
' G3 X5 M5 b4 D6 t% V5 V1 V, ~- |; v$ q
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:
$ V! \4 b6 `! a7 A. m! j4 i$ a这样算混淆了微分和导数的概念。5 x5 O$ l+ R1 m( u

. S2 K) t; ^% `+ V1 Sd(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
- Z. E4 j" }4 Q+ v! Z! {
, z& p3 r9 i, Q( g2 _' \Ah, you made a mistake: it should be:
+ W. x2 \+ Z, g% K
2 a& P' X9 S8 ~6 ?$ T x# ed(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:
4 r: ~8 E% B. RYes, there should not be " ' " after "f" and "g", it is a typo. :D$ U- \5 U7 N R: j% w& O& ?0 [

0 ~6 G y7 r( \1 @1 X: l* i, k% zBut 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:6 p! N: t. L" `6 B

8 x8 a5 {1 Y Q% d$ k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。
6 D. X1 X! ^+ B2 G7 k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:, g! O3 z9 j. i& X0 H8 U! @$ W
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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