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:* O5 [! I7 A; y" X( L
请教大家一道微分题，多谢！6 v. d( E# ~! ^, `- \5 k+ ~9 T

! s+ n5 H$ j1 c! c/ V+ W! ]3 cd [(a+bx) ] /dx = -kc +s 9 y5 h) D  t, B2 D
where: only x and c are unknown, others are all known,  requiire c = ?
0 ~1 g0 W& Q  M! x
  n: T$ x, y8 V( w; d8 `0 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:6 w2 Q/ r. @. S! E4 k3 y4 ?2 S
it is too easy:; |. f* {) u5 E" b. l: o

8 W3 S) `# s6 X0 r2 X$ ~d(a+bx)/dx = b
8 J$ e' D2 J) b0 [9 K0 B8 a2 Y7 [+ m
% y' P& X3 R5 l+ n) P- Xso5 t% `) S) X% [- O: R
! ?" ?: }4 X5 W: }
b=-kc+s6 L3 u2 S! D6 n3 [
9 d! z7 m7 g) U
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:+ e' R$ ^% F8 }0 N, e
请教大家一道微分题，多谢！7 A# L& `5 h5 G8 L2 U

( o0 j+ D( M" t yd [(a+bx) ] /dx = -kc +s
. P: F& c" G( O! iwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?- F$ r" F+ R {& w

( y$ [5 J" w' D8 J1 b/ G0 j6 O多谢了！
' [, _; Z* S8 B, c& O7 r. l* }4 E! `- d0 `+ I$ @& ]5 i
[ 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:
6 {. l& ?) j3 d# ]% \sorry, did you post another question? your statement is still not clear.
& j$ {3 ^# ^8 l( |: M! v
0 x; a% `; Q$ t7 C. ~1 e0 oa, b, s, k are constants? c means c(x) ?1 ~( M" G2 a: q4 u* A

5 L- n0 b8 v* R8 S( }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 12:30:9 k) R, w3 Z( o! n2 W6 E
对不起，写错了，忘了变量c,应该是+ S# s$ j' _9 ^$ w4 W7 ?; @/ ]" V/ M
d [(a+bx) c] /dx = -kc +s ) r( d: _; y$ @0 b/ y
+ g0 N- _/ E+ q2 T2 ?: _/ X1 b) 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:
8 c- c$ C0 A) ^% Gdo you mean it is:
/ l# g& g0 K8 x, i( i* K; E6 S
  W7 a! I3 H. _3 h% J9 sd { (a+bx) C(x)}/dx = -k C(x) + s' y* C& x5 M, [0 c

' O  j/ I9 l; R+ Hwhere a, b, s are constants, you want to know what is C(x) ?
; e  ]! [; h) i, ]8 T
* F$ ^: i1 Y0 t) Z* z" _7 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 14:33:3 {, F1 g& B- Q
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:
% V  _& Z# Q$ M" s  z% C8 T7 h9楼的答案有点问题吧。
. L, c% ]0 Q7 e9 g- D+ p) U  [6 {9 y您说：
$ J; N+ G4 ~7 N8 g
. y/ D3 R4 c. T5 E7 Ud{(a+bx)*C(x)}/dx =-k C(x) + s3 Q. h9 `% m$ H5 Q. |
so:
! J7 k- V# A+ X9 e4 m' abC(x) + (a+bx) dC(x)/dx  = -kC(x) +s8 H7 a& ^! `- s& }8 L' ?- g

6 e+ ]* W* u- _% t1 T5 M6 N  [/ k也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
7 b/ X% a( x% }( N% n% ~但这是不正确的
; R, W" w, j$ _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& H# s; U0 x! l5 K" R; S0 L: U
....! i# J& T ^# K9 X" V5 ^
d{(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:+ F! C6 E  ?1 k& o  s/ \
这样算混淆了微分和导数的概念。
0 F/ W. S+ j0 k+ k8 Q+ f
5 S1 c! ?8 h' e. i) q  d" @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
$ x) z& B) f  r$ S8 ^7 B) _* G" i0 Z# V. z2 y. _4 l+ y4 u
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 y" ~( }7 s( x3 z$ r这样算混淆了微分和导数的概念。
" K E5 e$ B+ O1 J* S9 P) {7 g, ^- P# F3 O4 J* G9 h* Y0 t
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4 l; h9 q0 H: J; I" z

) ^7 ]. i$ v# i9 B4 F/ b- D' yAh, you made a mistake: it should be:
. h7 @, b' ^( \& i: T$ a+ S1 A2 k4 Q
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:
) V) |9 d! I4 Q$ A6 H4 W0 wYes, there should not be " ' " after "f" and "g", it is a typo. :D
8 A# I( H! S+ @' t) q0 O( U: x/ B6 v' H
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:0 K3 W9 n9 Y+ q3 H% a3 C

) |5 l5 w3 U4 m% j$ m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。
3 L+ R6 }; ^( i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:
0 `. o% k; f- a, I4 i' C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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