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:
& W- i9 a) K  a9 p1 I: Z" p7 {2 C请教大家一道微分题，多谢！
+ P, A% H! t' \: n- V  z# ~: b: Z  c& N
d [(a+bx) ] /dx = -kc +s $ e, x$ }9 s, S% z$ n
where: only x and c are unknown, others are all known,  requiire c = ?) v0 v" I* ~0 o4 l, z# z

+ h' x7 a$ i  R4 n7 Q# y. V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:
' k6 j( H$ s1 |6 l  Y. E7 jit is too easy:& V& ~9 G& ~, J# m% Q9 {- e

& d5 @0 h/ [+ G; o! w2 Y$ Fd(a+bx)/dx = b5 X; Q/ n- x; ~' x  }; K, O. w
8 z( h$ C+ o: W1 B( o5 e: p
so
9 t: ]1 p5 }) l
8 y1 @  r; A' Db=-kc+s  e% U/ `: i+ O" D
- H. T( u8 _; A3 O8 d
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:
# w$ c6 G; h6 t% Q( u' ^0 B请教大家一道微分题，多谢！# v7 @9 M- }' `9 t( p. R3 |2 D
. W% s3 Z- }" q2 ?
d [(a+bx) ] /dx = -kc +s
1 ]* a' O1 i, _where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?# I6 e: q7 K6 W7 q1 [! ]
) ~- C) s9 d- U6 B. S* B
多谢了！
8 Y, e9 I- t0 \
; @+ e* }* |6 n' Q[ 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:
+ I. q: P* S4 Q/ v7 K) j% r* Qsorry, did you post another question? your statement is still not clear.
1 `9 L2 @/ ?5 n0 {) y( n8 n
2 U; ^4 N* j6 X+ `/ j. C$ da, b, s, k are constants? c means c(x) ?5 x; @3 m7 b2 Y* o" {: M2 m4 v3 Z
1 E# R3 O, e* ~$ ^( G$ l2 |, F; p
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 [( v' y* _' e7 _0 _
对不起，写错了，忘了变量c,应该是
+ ^# h2 B' W) U2 W* `4 _8 p& s( Ld [(a+bx) c] /dx = -kc +s
! N) e+ q! c: d$ H8 s8 ?1 @" v9 U' C- ?0 a! I0 z4 f+ J
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:  f  M0 t+ K/ T$ @$ g8 }6 V
do you mean it is:* |6 K. s. c6 ]1 f# e
( C% j1 n# e( i) N
d { (a+bx) C(x)}/dx = -k C(x) + s
# k* L" X! M- u0 [  P) z3 x% u
. t* o5 W, T  T6 a. m! J. owhere a, b, s are constants, you want to know what is C(x) ?; v9 a' {4 S7 C1 C' o: ~
3 A  L  w& `$ a
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:
I6 l" G6 w: j0 q* D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:% }- i% R' Q) K: a+ w% {) J. [9 s
9楼的答案有点问题吧。
- X6 x- m$ K8 l$ p2 I' h0 b您说：: u3 T/ P: Q( G, \
, V5 k; L+ z, k
d{(a+bx)*C(x)}/dx =-k C(x) + s1 r+ U3 s. H3 { _+ V; t
so:
6 T0 f! U O1 g5 K/ K; h. [7 MbC(x) + (a+bx) dC(x)/dx = -kC(x) +s6 S. L; I! b* p! E# l4 G0 \
7 ]( q2 `0 U& r, ?. c7 ~$ [$ v4 h
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
+ \. \) P! L8 N5 ~+ U但这是不正确的/ Z5 g5 D! c0 r- s
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& J; @# y0 q+ d( ^& `5 i7 T, X' D
....
' [8 d* ?8 O/ U3 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:) @2 A6 ?0 H5 n7 E) N' z1 p
这样算混淆了微分和导数的概念。' q, J$ F! g L B6 j) I

1 l; S9 ]) U, T$ Ed(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
. }/ L2 {9 Y5 Z' v' Z$ H2 @8 M% I* z$ O7 V) s, 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:2 ~8 V1 X' m0 z, g) _
这样算混淆了微分和导数的概念。
' |9 e0 J8 e) C: W9 f2 q: L2 ~% h0 p1 M
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
, g/ D1 W% A1 O' e& M: C9 N1 h1 ]* H W( N8 ?
Ah, you made a mistake: it should be:
0 F3 Y8 \- }" {$ T; B6 h; Z/ }$ L/ W( Z; H+ L5 m+ n
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:
" h! d" e, N. W- m7 Q; r' @& HYes, there should not be " ' " after "f" and "g", it is a typo. :D8 f5 x! _4 H w; `5 T

. U8 p9 M; ` c4 |3 @( ABut 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:
8 Y9 C6 \- R3 y& V: B1 c& B( V0 f% m% E* E2 |% |4 ?
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 d( Z. n1 r, S2 S: ~% I8 z7 e 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:
/ K# a$ f9 D; b$ H( 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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