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:
6 D* ?5 f1 P4 N2 X) y请教大家一道微分题，多谢！
. n$ q; d. w) U5 f0 t2 U7 b% j; D+ I: W2 i
d [(a+bx) ] /dx = -kc +s
- ?2 m9 T% h& k+ I* w# s6 O twhere: only x and c are unknown, others are all known, requiire c = ?5 V9 F% x- B4 o3 t2 o; h! \

( |" ^: {7 {, n0 v1 J' W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:0 e9 }, I& C9 p+ ]9 r7 G: o" q" K
it is too easy:% @' u1 u& ]( B8 v O7 q9 }& l

% {( g7 o1 S6 R! Dd(a+bx)/dx = b, N' D. H, a5 {# |2 q
- \: K5 ?- d' r* [1 j' b
so
, _% t) M" M( [! z6 l6 J/ Y( v, f8 U
b=-kc+s- A- v1 y' q: M
; R+ v( a& ]5 M$ i6 B5 B1 r
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:
+ b5 N1 l) j) y  l3 N' m请教大家一道微分题，多谢！
  B' H" s) |/ J- \/ l4 G
d [(a+bx) ] /dx = -kc +s ) r7 g, M& r# ^' P' E
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
7 B5 y' J  e7 O8 Y7 }" |. f) n# i5 n4 l- @. t- G0 [$ j
多谢了！+ s. D4 d; h7 W1 P
8 K* _2 n3 }  V: Y7 P: p- j
[ 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:8 A6 P  d3 b9 A6 ~
sorry, did you post another question? your statement is still not clear.2 \  Z+ p( K, N. m

/ d7 `& d7 h: W, U4 Oa, b, s, k are  constants? c means c(x) ?
. R$ N+ i5 r( ]; b: t/ h6 H: q' v) a4 M; {7 N
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:% n7 C4 K# N* e+ R* g: m$ d4 j( L
对不起，写错了，忘了变量c,应该是
8 ]1 D8 ?: V: K$ S6 xd [(a+bx) c] /dx = -kc +s
% l, Z& z2 |6 _3 T B, z- `: L: A3 @0 x1 e, Z4 k e& X
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:
8 K/ n' S4 l+ }4 p! s. w/ t. E& j# Z7 Jdo you mean it is:1 D3 `: Y& c2 [. |/ x' M
; ^) e7 x$ Q/ h+ f7 z4 h, [# |
d { (a+bx) C(x)}/dx = -k C(x) + s
& E1 i, _1 N- |* I) P$ I2 f8 M. S5 Z' `3 r2 a9 M. b
where a, b, s are constants, you want to know what is C(x) ?
$ f- @, i, t. ~) e: g. X6 N4 B2 d% b; Q$ V' q4 w" v7 Z
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: W2 P5 u* @! ~# Z4 [
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:- [1 U* ?* \: B; h
9楼的答案有点问题吧。
% |3 j% E$ J" I; h: b3 q3 W您说：& U8 E+ @* p7 r+ h+ S& k! Y" Y

' V) W- U3 _4 p1 T9 y; ld{(a+bx)*C(x)}/dx =-k C(x) + s7 m. e3 z# K! X$ \) Q
so:! f" Y0 S/ B4 B5 d% }7 p
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
3 x8 I9 D: _, Z& n3 ?3 Q! C% B- ~4 b1 V
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
* w- B# z/ v$ T! t" N但这是不正确的" |0 w3 s6 K. X {2 N
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
0 y: S* v9 [1 I! Y) T....
$ H% o N! s2 Z' n1 td{(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 C9 z2 c! e* c' ?
这样算混淆了微分和导数的概念。( }/ `( I8 G0 Q( O* v
2 J% k* E$ b$ r6 ~* p
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: z- V6 `6 m- d7 }
' o- n, b- p" f9 l! p9 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:
7 V) Z# a0 i" {$ p这样算混淆了微分和导数的概念。) f' M2 v7 R! F/ Z% @

2 p6 w( D+ K. _ a ?2 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
" L( F, p) z% ^ K* u# c& c6 @9 p {; _8 d' P# T4 {; r( Q
Ah, you made a mistake: it should be:1 Q& r: q9 O8 |& D; b9 F

- V) }) q, \, W2 g( ?3 |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:
! @6 U4 X. g$ U$ B( k" vYes, there should not be " ' " after "f" and "g", it is a typo. :D
' O2 p+ u5 U1 J3 _ w
! l; ~% Z- K: r6 Z; O' hBut 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:+ l! |4 d* L, z, R7 g7 u6 _

8 u% ]! v. z/ D/ f% A7 q3 ]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。
5 T6 ~' t9 M) t) B* ^2 D* 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:
( U; Z. }- A4 O4 F/ k9 R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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