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:; q) H- [& H0 p) C$ H, g# Y6 z
请教大家一道微分题，多谢！4 h# R" |) Q+ J% \
$ c) ~: m- O& m3 r3 a' i
d [(a+bx) ] /dx = -kc +s
" P3 b1 P, p! R3 wwhere: only x and c are unknown, others are all known, requiire c = ?
0 v* A* X) |3 `7 U( ~3 m7 z- I4 A4 n
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:
( W" M7 C& X- z3 ~) U" k) q( Zit is too easy:
6 H& m4 i8 m! n4 c
8 f+ K6 d, m1 V# x2 ud(a+bx)/dx = b
  v4 J3 }' h. I
" H/ n6 Y% n4 r$ G; yso8 B& H7 i& t. C/ F: T+ S
7 E" {8 C% l% u  j
b=-kc+s
8 V- t9 i& n& {1 f* G- M5 e
6 l; ^  m& f# s8 Hfinally:  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:' q% L3 d3 p9 H/ \! j: `( |
请教大家一道微分题，多谢！
$ F0 E4 ]2 j( g: y( X* }
, u1 ^0 ^) b$ c+ }! y- zd [(a+bx) ] /dx = -kc +s
/ J. d" O. K$ Z4 uwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
1 s: L' u) s. D7 G4 G4 z [: b# M# M' ]6 ^4 T# a
多谢了！) o3 y( w) r( \

# L9 g. C( }. D. O$ ?7 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:7 l9 p% a7 S1 `, G4 i
sorry, did you post another question? your statement is still not clear.8 J" p8 {# y! |( g1 F. a) O- m0 h' j

+ O- H0 }# ~6 r' F+ T9 t* ~+ ma, b, s, k are constants? c means c(x) ?
0 F/ s+ F4 y/ K+ n8 B- \# g# V4 k+ X0 M! c! O+ g
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:
/ p8 m3 u2 @; |4 `! |9 c对不起，写错了，忘了变量c,应该是
7 w$ a. r) V x6 _d [(a+bx) c] /dx = -kc +s
+ M/ @5 D9 i3 ?5 o3 Z7 w& t* |
9 w' i+ k$ }# n* A' ~6 w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:
5 I: ]: E1 U: `- ?* S7 \* W0 ldo you mean it is:
( l9 ^1 F/ w9 z6 p% d% `6 j2 U1 R9 f& u8 E" L
d { (a+bx) C(x)}/dx = -k C(x) + s$ I* q/ N7 T# y9 f4 e C, T9 ^8 x
' b0 K/ P1 o7 H# C2 u
where a, b, s are constants, you want to know what is C(x) ?% D/ z8 U& q3 r' \, r+ K$ N

) S8 A$ S( S" 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:
5 i7 q1 r0 u* M) D, l- ~; e0 @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:
/ J5 M8 K6 Z. I9 c9楼的答案有点问题吧。
您说：
* D6 w& s, c3 E" _. _
+ d, R t3 K F9 E$ \d{(a+bx)*C(x)}/dx =-k C(x) + s
, F1 V4 o6 e6 A4 |3 c5 H. q2 z1 zso:
" P! z) u8 p/ q2 s# XbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
# E/ u* J) \+ U: F4 r, Z( d* r" v, b
4 W; Y+ G9 b0 |也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx1 E) r7 N* y. Y& O: J: K
但这是不正确的6 `' E" S7 ~+ ~: ?/ Y" 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
: }* F( g, N( w a$ |" U- O& j; k& `....
. s' B9 i; S" Ld{(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:
/ r$ i* T: @' W0 n这样算混淆了微分和导数的概念。1 k4 s9 p: j* s( x# J

0 Q w; r, p* \, h0 Od(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
0 \ r1 H" {) U b5 M( b: `+ _% [, T+ `
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:
& s$ g7 v. V' o& i; C4 c/ n这样算混淆了微分和导数的概念。
. r5 [& j2 ~% Y$ s+ t+ o; O$ g# x5 Z
, `1 G m+ g6 I+ {2 r& F% T& Td(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
! W; N0 T* \! l" u+ ^- P" D! h
Ah, you made a mistake: it should be:6 ?2 q1 j3 d0 y, z; w S) c
2 P) A6 q4 M# h
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:
2 U0 b( t& e5 O' z4 hYes, there should not be " ' " after "f" and "g", it is a typo. :D* _ K2 X: f- M& r
; A H. p1 y/ {7 o2 J: y6 X, w) m' u
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:
( }8 [6 N2 f9 t+ J |% b3 c$ ^1 c
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。) a4 S% X+ |8 u, e# H. r) |
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:
+ W3 m* k& b* K. r/ t$ e* ]( {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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