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:
( b: @# f1 w' w- W请教大家一道微分题，多谢！
$ Y" ?7 e) v5 q: H+ I
/ s3 w1 Q$ m, u: l9 C) h9 ~9 Y* kd [(a+bx) ] /dx = -kc +s ' m$ F! D3 S# z7 s6 q1 v, z6 T
where: only x and c are unknown, others are all known, requiire c = ?3 F' p- M B9 M; j9 k5 m3 Z
/ a5 s9 |) T( D W8 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:5 F3 P9 L+ l/ @
it is too easy:
0 j  e& f" B2 t0 o
3 B) e; T% |1 B) S1 |d(a+bx)/dx = b/ J; o6 f2 m2 [9 S/ Q3 t+ B2 F

- t$ [- ^3 i' C2 `1 R; p% p, ?% Lso
3 ?* d0 \5 V$ k
6 d" P& `1 @: n/ z/ a1 Bb=-kc+s
& J  t+ v) f' U
5 z6 b" S5 F1 `2 T+ Z, K  vfinally:  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:
' ?: F- P! B- H* [3 B; _. M请教大家一道微分题，多谢！. G. R& q( c$ h
" P( q/ Z+ `$ r! ~! f/ Y5 v
d [(a+bx) ] /dx = -kc +s
3 y6 [3 r( S2 A# c* dwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?. S0 }5 a! I" O6 y
7 E2 t! s2 T# g' P$ ^% Z
多谢了！' Y. R2 ^/ d. J& X1 ~# P
" y) b9 ^% Y/ ?2 S( {, Q0 R
[ 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:
4 C7 x/ ^5 U- d9 d/ psorry, did you post another question? your statement is still not clear.
# X/ S; a) r$ f! J/ U1 o: ^5 k0 X  _8 l; p
a, b, s, k are  constants? c means c(x) ?  E. ^+ Z, N, \! @/ T
- M6 C$ y1 O0 ~& U  Y
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 12:30:8 Q( i5 P* U' p# M' A" X0 |
对不起，写错了，忘了变量c,应该是
+ Z. n9 y8 [ ~" m- q) I9 Qd [(a+bx) c] /dx = -kc +s ' t- O6 r" L( i1 Y

9 E, Z1 n5 ]. Z: q5 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:* B4 y3 l+ O& o5 q0 L: e
do you mean it is:
& |' C. a, [3 u7 e* W' {5 Q' X
5 P2 d! B& k5 vd { (a+bx) C(x)}/dx = -k C(x) + s
& X8 V6 p E# m9 Z6 M
/ i) i1 d/ J! C. y" C/ ]where a, b, s are constants, you want to know what is C(x) ?' d4 ~# ]$ s3 U4 e- \1 H, I

) ?* q& _8 d/ |! H3 u! _5 mif 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:/ ~& F) N/ K( T) ]" v1 X! S$ U
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:
: E' a' c0 U9 [9 R+ k! ]9楼的答案有点问题吧。
: z( _% S+ w* L- b您说：' n. A: ^% e" ^' `! N% H/ H! v
! `2 a- V+ c, [# X- i' T
d{(a+bx)*C(x)}/dx =-k C(x) + s5 y" s- L% ?6 Q) K" I
so:
5 U/ G3 |8 T- g* ^; ^bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
& `. ]) R! z5 G; V4 a& F3 c7 U3 x9 h$ R5 V7 |$ ^6 _5 D
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx- I7 j+ F/ F! }! Y7 D, ~, L( G& f
但这是不正确的3 E4 N; M( r# {
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
: @( a/ F* V7 C2 C; Z....
' G+ V# r- n0 I) v9 u& L [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:
1 {$ v# r: N" v; B, c2 ^% C7 v7 F这样算混淆了微分和导数的概念。1 i. r0 X- C2 V4 J

' t3 @# c6 @0 v2 L9 I4 ~; Xd(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2 n) W( G% R8 w% A( W* R) z

+ c3 }" O9 S& ^+ w( _5 mIf 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:
$ j$ C* c4 U) v5 t/ X! r这样算混淆了微分和导数的概念。; G# G8 o( }% n9 @5 z$ e, M2 c# S0 j

/ H' e( L+ q6 ~ id(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$ R' m9 @8 s/ N# X* K
# B) Y3 C4 c7 `& T
Ah, you made a mistake: it should be:
9 k3 o3 [2 y" f" _! N
% I3 B _# E8 ]; B0 @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:# M( {+ w m0 S. G% w
Yes, there should not be " ' " after "f" and "g", it is a typo. :D( G3 v; S$ L# ]* K. I

( _$ t6 @; Q Z2 DBut 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 i+ |7 \0 G' o2 a) e* o

5 s( W" l o( 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。
- h% Q6 l. K% p' Q& E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:
; |% L! { z8 c; A4 J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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