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:
" k2 X& L/ |( k# a6 u" z8 N请教大家一道微分题，多谢！
z8 O, y$ z( `& q3 ^) a z$ [- k$ h6 n
d [(a+bx) ] /dx = -kc +s
0 i. X: y& M Z& mwhere: only x and c are unknown, others are all known, requiire c = ?3 _5 O! k, x' M, y. C

# e7 t, f& X5 r/ }/ u! m5 i& r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:. H% G* n+ W. y- A. Y
it is too easy:
: T) N  J" x/ z7 Z1 j' d- \, D3 ~+ _# `  ?
d(a+bx)/dx = b. g" ]' U% v+ d. V3 W) _0 U% D
; L6 H  W2 ]+ {* N
so5 z9 j, ]( a7 }1 w; k" s" V

/ t/ y$ ~7 J$ D$ _. y# Lb=-kc+s3 }' y1 |; D+ T6 i2 F

8 p; U* Q+ v5 P  s: H0 Ufinally:  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:
* `- P6 M: i. O% |& A' I- w8 Z4 d请教大家一道微分题，多谢！
! e! i/ J" e7 B" e$ l- z3 X& T, ?1 }, B1 ?$ z
d [(a+bx) ] /dx = -kc +s ( u& J+ Q/ G( w) J6 {
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
; d6 Q; A- O+ U3 x8 r" ~$ p. |
1 }* y1 Q& Y6 [4 c* ` E% _ |' A多谢了！
/ w* A& j: Z' \2 t# ~1 Y8 g! S4 b; D
[ 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:6 t) X0 w. B1 w  l$ t. f
sorry, did you post another question? your statement is still not clear.
; f5 t5 h5 h4 V  n8 v2 g" R7 X& A% i  e& e) l( h
a, b, s, k are  constants? c means c(x) ?
. w/ W* B5 K4 c. F/ k5 v) G: A6 k6 A+ D& I9 Z- E
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:" a0 R" b: h. b. W( f$ d+ z: Z
对不起，写错了，忘了变量c,应该是" e J) V( U! y" ~& `
d [(a+bx) c] /dx = -kc +s / G( C- Q2 i: Y: }5 D

J5 l) q0 \2 {8 ]$ E( b) W7 Q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:
0 a8 ~( m: W& N/ v$ F* ido you mean it is:* y) d. Y+ T( l. Z: Y# }! M) \
, K$ }8 t' H, j. ~! ]1 G7 o5 f7 X7 H
d { (a+bx) C(x)}/dx = -k C(x) + s
. N7 g* m# l5 ?5 P. a. ~7 U p7 u2 Z
where a, b, s are constants, you want to know what is C(x) ?
7 Z8 ^0 e( l; o# j
5 O' u7 @7 ?6 l( }+ i3 gif 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:
! d$ \3 n" q x9 `6 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:  ?& a( g+ [9 Z( ~: X: W+ c$ |
9楼的答案有点问题吧。" m' {1 E: h' |- O; \! i
您说：& g" f2 a4 B* m% \( T( p9 |

3 r) d+ o# O/ B1 l" \- p& zd{(a+bx)*C(x)}/dx =-k C(x) + s+ g8 T* d4 b8 H3 c
so:# ^! F. X, R1 B4 S% j# Y: j
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s3 [; q+ C) P( g$ D$ u' F" M1 t

, s5 L  J) R2 E也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx2 b2 e' g. t" M3 D4 e" B
但这是不正确的" j# x4 r, ~7 w; }% l
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% d5 M/ L! A& w
....! t; K6 D W+ j# d7 M* f" {
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:
0 r+ M& J7 V3 m% g5 E这样算混淆了微分和导数的概念。
1 l0 M9 m/ I ]- ^2 {! Y5 q) m
7 t E+ e4 o6 F7 dd(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 |, |1 R- m1 X/ E* M; I/ z1 X9 V
) M* }+ T3 A" u9 j$ u3 cIf 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 f+ N, W# g; o+ }& G2 R
这样算混淆了微分和导数的概念。
7 C/ |/ S h5 r( O4 t# z9 S/ k& M4 B* ]) T9 r
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
A- Q6 Z7 e: ~% c8 }: v5 L6 ~! R& J/ [3 g- _* b c; M+ c0 f
Ah, you made a mistake: it should be:9 z. N; O, ]& k

" q/ p8 ^- U. r" gd(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:
; k: U1 v* V) {% yYes, there should not be " ' " after "f" and "g", it is a typo. :D
% r! ]& B' a% P' X3 m }$ \
$ w" P9 v, O( x1 w" E/ x" uBut 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:
' k0 j$ T' X8 m* w |- ]7 W7 ?! i+ n0 i2 q4 W3 `
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。+ G7 C0 ?4 ^6 n- Q: t2 P, `) n
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:$ T/ d# w. Y2 L _4 F- }4 Q3 y
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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