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:/ n' j0 X$ y( @0 ~
请教大家一道微分题，多谢！
& x, b2 V ]2 g, s- r: u3 \# \3 M
% G; e( Q: J, ~+ i% nd [(a+bx) ] /dx = -kc +s
: W" Z) M9 m0 f& R' G- S4 _) bwhere: only x and c are unknown, others are all known, requiire c = ?8 q- Y! h! c5 u7 f4 I
' m5 c! Y/ E6 l# ?6 u( z- _
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:  Z9 q. y, Q2 c% f) d
it is too easy:( v# ^" m6 G" \* X* m4 U

& Q; t0 z9 m) `, F1 d8 |' U& ud(a+bx)/dx = b. K) l1 r  F+ A. ~  S* I( D# N

5 _  N  t" I$ Z% t" Cso* s& o4 c# {9 e+ ]( c: V- x

8 W9 ?5 a/ K$ h+ l5 g8 f" J* l2 Xb=-kc+s
. b5 s  U1 I4 b& w& _. s- u. f, K) c' Q
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:
/ o0 T4 s3 n& a/ O8 q: u请教大家一道微分题，多谢！: q3 A& j8 q) c2 \
1 P  v0 ^* f- t4 D1 a! t
d [(a+bx) ] /dx = -kc +s 5 U+ B0 N* Y8 z. s- {0 Y: M4 F( ]
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
4 c, o1 t  F; h- X$ e
, S+ T# j1 Q1 D) @& e& R+ _多谢了！
- F* I  a; }1 v
: V9 P0 g6 Q  p5 V. P, L' P! ^[ 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:
0 ]- y+ m9 L$ K7 |& o2 ?sorry, did you post another question? your statement is still not clear.! r( A6 p/ \1 P: p( D( I7 _% P+ n

+ ^. j1 A% H" Ya, b, s, k are constants? c means c(x) ?
8 k0 H) V1 T4 ^6 g4 J# [9 m' K
: \1 \! D, v/ f% splease 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 12:30:8 e( X$ n* Y5 V9 e: B4 b z
对不起，写错了，忘了变量c,应该是' o% w! E% _0 F
d [(a+bx) c] /dx = -kc +s 0 b$ D' ?& p; V- |/ Q

$ ~0 A* ~- d& P3 }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' F0 k) V+ x
do you mean it is:6 X5 p0 ?/ y. x1 Y7 v7 d, n( g

1 z: V6 D8 o/ K3 kd { (a+bx) C(x)}/dx = -k C(x) + s) ~: h+ P* N7 c4 v
. X, Q% X* _; o- w: f* _$ A+ |
where a, b, s are constants, you want to know what is C(x) ?
& N+ C' r" L! n; D1 D- N6 i
1 [, r0 A+ z' s' A1 F# Xif 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 14:33:0 d7 x! B4 t2 r9 u, x' u2 {( r& n
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:# v7 |7 M. J: d! e' s& n- ^
9楼的答案有点问题吧。
; `8 S6 u) W/ ^您说：- A5 {! }- l3 l# Y4 @% F
4 Q6 E/ ~& C- h5 i  {! n
d{(a+bx)*C(x)}/dx =-k C(x) + s
; v9 j6 M/ n/ w5 {6 Zso:  V8 \6 g$ }* F& \. _
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
& v; h8 g6 \9 w$ ~. G9 X
8 v9 P9 _4 e$ k/ Z  L  `也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx+ S2 |* K, @- [2 i$ \4 Y( c+ X* M
但这是不正确的
4 m0 b/ R* o9 Y4 ~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
/ T7 r; A+ `$ ]6 w....
9 \ B; `, u8 vd{(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:
6 o& k$ d; r& N) o* S6 N! l$ A+ k; |这样算混淆了微分和导数的概念。
' F* q h. Y8 ]6 P. V W# v9 w6 r5 Y$ \7 j: ^* U$ E
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& F4 s% t. s7 S+ m. m- Z8 }" J- Q
7 ]( j! M% B5 a! b/ W! G. o
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:
这样算混淆了微分和导数的概念。8 n' |3 b7 |. D
8 b& S1 a+ ^$ N h8 U/ [
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
7 T. D; H8 B0 X4 c7 ~% X
. T% `, \+ G. e/ K1 j, DAh, you made a mistake: it should be:
, r) w0 Z* S0 G4 h
9 D& S7 r3 [+ n- G% J: md(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:" n6 V: g- Z* E, d
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
5 S( k6 B/ G( h2 c: i/ E% a* |; C9 w1 x" I# 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:
4 B6 B o0 Y+ ?$ @% V* l$ |& W
" H7 T4 b& C/ P5 ?9 L& d2 d" T' }/ `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。 C8 W/ g8 b6 h* \+ M4 W6 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:
6 n& D" u8 J) H# V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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