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:
* r4 G4 P) o& c请教大家一道微分题，多谢！
& Y) K4 a3 m, ~4 u6 }+ ]0 x' `' Y6 O
d [(a+bx) ] /dx = -kc +s 3 \* n3 M) O& \  b- E6 L
where: only x and c are unknown, others are all known,  requiire c = ?
  d4 C/ I0 }( E; e# A$ z; e, ~/ y* ~4 _0 s# n$ Y
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:* g$ e: M) I  G
it is too easy:" q& H2 Q6 X' u8 B( r

; q3 N; I; ]* cd(a+bx)/dx = b
7 T0 ?1 i( o( t, v# ~' r/ v
  s$ K9 R# D8 ?6 m8 Kso' F" y  k8 V3 \

- D. V4 C( @' O; Wb=-kc+s: @2 h& D/ e: @& c7 x
$ E; w; ]) [& f7 m( z* J
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:
6 `! M- d: j) I% {0 \  \请教大家一道微分题，多谢！
; O7 w1 S1 l% `! G- ~; U* d( v& Z7 t
d [(a+bx) ] /dx = -kc +s
4 Z" ?' {' Y9 l. awhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
9 y& V9 x* E5 b- o/ h3 M1 n
( v* {* u  `: G5 h多谢了！( C" [5 u/ ^' N. q; _
) g" ?8 i& l: ~' L" p, }0 v
[ 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:9 x. Y% z7 I3 }! f: D/ Z
sorry, did you post another question? your statement is still not clear.' S1 X* J/ O1 z% f" {$ }
) B( d+ j( ?! a( c8 }3 b
a, b, s, k are constants? c means c(x) ?
3 B8 O* B$ }0 x' f4 Z
( ^ [8 {5 D' S. t( e/ 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:  d% q+ J" b9 C( H9 G3 o& D
对不起，写错了，忘了变量c,应该是/ U  Z# C1 q% Y8 `# J1 c
d [(a+bx) c] /dx = -kc +s 8 U; r7 a6 N* v* I: _  k

* l, r: l  G% U/ x: \1 v" ]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:( _9 w' z* [) P; A
do you mean it is:
( l6 x' r6 T6 _8 {& T' b Y+ H
; D2 \# q8 f. sd { (a+bx) C(x)}/dx = -k C(x) + s% @1 _, s2 ?7 f0 b1 a
4 G7 h. X/ _0 k" X1 o- [5 j( J- e
where a, b, s are constants, you want to know what is C(x) ?
8 b4 M, k. j' _ W. X; r1 T* F6 @/ l
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:
l0 ]3 `" M5 s& o# g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 19:34:9 A. y) \5 n0 o2 g* t1 G3 Y/ L
9楼的答案有点问题吧。
/ a; ~& F6 `% n  b您说：
" l2 B( M+ z1 F, k5 B" }2 r: b  P9 e  Y& y
d{(a+bx)*C(x)}/dx =-k C(x) + s
6 q' G' ^" L3 Y9 _- nso:
/ {: c; Y: v  h9 ?; PbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s% k1 @1 G* J5 ?" ~6 X

& ?5 F- c& M. w: _% L; A. A也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx# o7 ]% a+ e( s# L+ R
但这是不正确的; z1 i0 A8 @' J2 F
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: e* ^) t+ E9 x! C
....
# V, s/ G5 k* u$ ]9 H* p! Ud{(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:
% ~" O2 b% [4 S4 }1 u这样算混淆了微分和导数的概念。9 O& T* ]. J' L$ `

' d R6 z) b! I/ L$ h. S: T* S; Hd(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 p. d1 j% e- Z; V- F% C, }
8 E: Z, W" H7 I% V/ @% }+ b+ e
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:5 { |# }' X, W4 `
这样算混淆了微分和导数的概念。9 ?2 |& Y/ o. S( |6 e2 B* `6 _

. i9 w! O' p& r4 wd(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
6 c0 _: r3 |- o( c( h) ~1 h8 D- B
Ah, you made a mistake: it should be:
z, L: x, j1 g9 s q
) T3 K2 ]1 b/ T+ `% Hd(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:. @$ [+ `" q0 ~) e' ]
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
0 r Y; ?, n! q" }! g" p+ C8 E- S3 ~8 j5 t; e# L; ?: c( j
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:' @; X/ X. t9 ]( v* y

& L5 m: m7 F- ]' k/ U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。
! F% {' k" I3 g8 }3 |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 23:24:3 u" l( [' d8 O( p9 ~' h
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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