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:
0 i; r j% c, S* V1 }2 t) J请教大家一道微分题，多谢！$ k) W+ s1 f9 f, w# @% s% [
3 W# }. ?+ c' ?
d [(a+bx) ] /dx = -kc +s ! n" J% d4 n# J7 x% R6 P. Z
where: only x and c are unknown, others are all known, requiire c = ?; x. [% Q; q" a, w S3 u8 P$ [7 Q* a

$ f! t& e* ]+ ?: A. ^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:
2 Q' N& o4 W/ E" W0 D  _3 dit is too easy:1 `5 s7 b) g5 E/ l

. @% d% @) ?  {+ }4 `( Kd(a+bx)/dx = b* ?6 q# j2 L2 Q5 a' B) k

- O3 e/ }; x4 m% xso
( x0 X% @1 E, o8 A( n' [
, g. M6 |9 H$ O8 G. x% Ib=-kc+s6 N0 p! g7 r& z6 [; j
) ?' l/ O  o% 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:
3 |1 P# G* E1 n7 O: M0 l: d" \请教大家一道微分题，多谢！
; _0 V, ]' H! q( ^. Y
% _" E6 b) j7 |d [(a+bx) ] /dx = -kc +s + W, b3 V1 x" [; J, a6 g# U
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?! m" h+ v+ `6 i8 z8 C* F) H1 q

% p2 C5 a* e6 V. s多谢了！
1 Q- \, [# _) y. H' _! `/ r1 E% N k+ G+ |: }7 p1 y W
[ 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:
1 k: \2 C& q8 M! ^% ksorry, did you post another question? your statement is still not clear.6 P. w; j2 c3 b8 |2 B
$ T# N! P) o% f4 q, b
a, b, s, k are constants? c means c(x) ?
) k. \% z. C* T& i# v P
9 c1 d1 y: G& U4 D1 @/ y" hplease 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:
" g) S" c# ^/ i: _对不起，写错了，忘了变量c,应该是
# c: ~- ]9 T6 s# M" ]4 Kd [(a+bx) c] /dx = -kc +s
8 \; k+ V% M% f, G
5 E$ G$ `" G+ z4 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:( x- }9 s- h* u: G( c4 Q
do you mean it is:1 e% N; _0 E, G
R. _' y" A. q, T* E
d { (a+bx) C(x)}/dx = -k C(x) + s
+ B( _; E" Z! Q% E; m2 p5 J! O% B8 [, i5 k+ Y
where a, b, s are constants, you want to know what is C(x) ?
: N- v# Z* O8 c- `7 d) J5 [! ^+ k" ^
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:
8 n6 p) U, o b7 \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 19:34:5 F7 I$ \  R' e: @$ R% b: l
9楼的答案有点问题吧。
" j5 F5 O* @2 \" |您说：2 [8 r! t( M& K  X$ a5 C! U  w
+ H4 D8 m' O; s2 Q3 H2 B- k
d{(a+bx)*C(x)}/dx =-k C(x) + s
1 y  p" \! ]( u& T3 `so:
( L) h. U1 |) ~bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s4 M& X( F/ B* M/ i' [8 I

5 y* S- a. ^! h8 m# e也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx/ r# R1 i% m9 s) o% \" e. g
但这是不正确的
) j% Q+ B/ {/ G0 [' M2 L6 Zd{(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4 R* n8 n0 ]6 K$ `$ O2 A( Q
....
6 h% t5 ]. \5 ?# _( C4 L @, ?, Hd{(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:+ z. s$ O" \: @2 }; O
这样算混淆了微分和导数的概念。* \0 ?2 k I' ~& I8 t# `# [

6 U7 z) P' Q8 V* W: |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
/ b( n, `! ~& C. s' v
) `1 L% j; S+ ^0 q- \+ YIf 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:
" P6 z9 D' e& Z7 `% L4 D7 q( P- K这样算混淆了微分和导数的概念。
- f7 n) n; A0 n. n8 Q3 D
9 ~* M9 u( s8 b; P& 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
! x' T) t2 L% Y0 B" T; A/ o' j2 Z, h) L2 l# X$ c. f& U
Ah, you made a mistake: it should be:/ o: C: Q- w" n) H& o
: g" y$ T' [0 d# B x2 X7 i+ K
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 20:51:3 A; K. V! F- `. l, E9 p& x7 V: y
Yes, there should not be " ' " after "f" and "g", it is a typo. :D2 N% P. U/ ?" a5 I
( S2 t, w5 c9 `/ X9 T
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:1 I4 Z& L6 U& T1 P2 I. b( G
6 B2 U# V" t' t9 t) y
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。
' a [& E, A8 k4 a$ c1 a5 l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:
! `' Y( S( Z: E3 q1 c$ U" L' I. \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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