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:
; S1 y1 B& F1 }  Z3 h, o请教大家一道微分题，多谢！  q! W$ Y" f  I7 K

! n  v+ S& L; D7 @. F; dd [(a+bx) ] /dx = -kc +s 7 E3 p* a+ y2 ?+ h
where: only x and c are unknown, others are all known,  requiire c = ?
" w! n  U! I1 q9 K
( n; d, \4 h  u8 X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:
+ \& N3 V. E& rit is too easy:: D4 A# \9 E) q$ ^

* [- `( \' z9 e" H9 r2 w6 Xd(a+bx)/dx = b
; w+ m" i2 a) |1 d4 W9 z9 m
( x$ s2 L' o$ _9 t X. Dso% t2 V$ p/ e$ t+ q3 }$ B# P2 @

9 V/ n% o: g; C0 [b=-kc+s
+ t/ }% H2 h5 k2 K' v) d% n' D/ i& q- G$ g7 a
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 20:41:4 ?9 j) i/ X: a& A) g  r
请教大家一道微分题，多谢！
9 |( O2 s7 ?5 X2 d- `. _
- n- J% S7 v6 kd [(a+bx) ] /dx = -kc +s
; G, C6 o# V3 S0 c( F9 C1 }* R$ Hwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
, j0 ~- g4 W; O8 t9 m% W, {1 [  d% k6 Y
多谢了！) H, P  H/ X4 G3 e# m

. o+ A3 z- s' B+ V2 t7 p9 @[ 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:2 n; I1 ^9 F6 i' g2 o
sorry, did you post another question? your statement is still not clear.
/ B) F) X* n! i2 l: U; u8 V+ m3 |0 E k( [1 o; Q K2 I
a, b, s, k are constants? c means c(x) ?
9 y# v5 ]+ g2 G* N! y' b* o' j
4 v- n/ m$ Z3 [6 aplease 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:* [, J$ Z' q- H5 B9 I* N" `& l
对不起，写错了，忘了变量c,应该是 O, \7 g, r6 ]4 }! O* S
d [(a+bx) c] /dx = -kc +s
! _7 @! o5 \7 _0 L. Q
3 ~" y3 D5 ^2 @* z$ t4 g3 y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: J- v6 s$ s% K. ]: F7 H
do you mean it is:8 G( Y: h! F5 x4 }. p4 |

4 @# l% f1 \9 ad { (a+bx) C(x)}/dx = -k C(x) + s
% o, {: E: w6 p4 h% ]
1 X" w: O* I* O4 wwhere a, b, s are constants, you want to know what is C(x) ?% J# o. w5 M1 s8 T( N+ o3 N

6 L" B; x$ ?8 Z* Lif 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:
8 \/ t3 \# S1 k3 B1 U9 G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:, D- U; Y5 |: p
9楼的答案有点问题吧。6 W/ a# @  N: o( H6 ?
您说：
& T, \& L+ c# W' j: F; ?: F! [2 `. ~, n" z$ _) j
d{(a+bx)*C(x)}/dx =-k C(x) + s
) _2 T4 a; L9 a: c8 k/ t# t6 p1 u5 aso:
& B. `5 [' {7 p4 C' ebC(x) + (a+bx) dC(x)/dx  = -kC(x) +s5 F) R3 ~% q7 a1 i$ \# r4 b8 k
( Z. I  g1 O/ }3 v
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx# V3 z% O/ N/ t8 C. w  V
但这是不正确的
6 z" w4 V4 T- ?( z5 \! td{(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6 \ k5 R/ t/ w$ a9 `
....
% e) W" X6 t: ] k: fd{(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 E0 O6 \$ T9 T这样算混淆了微分和导数的概念。5 F y5 W' [! [# Z u' A! T

' S2 Y. Q7 _+ ^5 V; C0 j) Vd(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
2 j V) t; o$ W# i" A2 S" V0 c
* g) C2 E, H: t/ {' G3 ]% e, ^1 PIf 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:
1 Y" B$ ~2 L; K/ z这样算混淆了微分和导数的概念。
3 `$ D8 J1 a5 V5 n) { S3 s, H/ h. c# f2 f% m8 ^+ L
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
; H7 P% O" d' s2 _1 o/ O. t3 Z
$ M$ h* H3 X0 u3 G8 C- E7 FAh, you made a mistake: it should be:9 o, c* L j" ] m
" l% E% v, \; v- f+ r( c6 L
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:
9 _. z/ d/ ?0 i2 O0 q' w+ U' mYes, there should not be " ' " after "f" and "g", it is a typo. :D1 k2 u. b/ m2 T5 {( W
- z1 J/ T" g7 n+ G1 u4 O* ^ a( v
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:# a; E0 {# H( f4 Z: T z
8 L! L6 w: D4 n- |' H
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。$ A5 x* v( o7 u. G; X: U
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 23:24:7 b( T3 b8 _2 \, [+ ^
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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