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:
7 t& V6 a0 _ Q& E$ q) S请教大家一道微分题，多谢！ D7 X, T: v* f" a" A1 o
& i1 ]5 u% a. t1 s4 q1 E* ^$ V% D$ B
d [(a+bx) ] /dx = -kc +s " t! f/ W. |% T* c( s) e. P
where: only x and c are unknown, others are all known, requiire c = ?
0 N# |6 ^3 ]. h' R# x% F( O+ n0 I* H/ f% F; f5 o5 P9 u
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 z) J2 Z& e% X" M  |it is too easy:" N/ j/ Z7 R/ g$ j9 E8 L

( g/ P! d5 ^! g  v! Bd(a+bx)/dx = b9 [0 l2 N; G2 h

% e/ W! x3 L$ A: oso! \# ^1 P; Q  ?) }7 K  ~  Z& J
( R, ?9 [7 _( Z
b=-kc+s8 ]& r3 \* Q! Y  Z& u) U9 @* C
7 b7 ~  T' H/ p4 Y( \7 D6 A
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:
% u" I  J; R6 c. s0 }请教大家一道微分题，多谢！4 R$ ~1 w: Y6 |

4 U% r+ F% N! F  Qd [(a+bx) ] /dx = -kc +s
8 y. }# ~+ O2 a3 O9 vwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?4 p! P2 l, z7 W. k

) m/ z6 _' E& e. b多谢了！
' V- M# n$ Y' v+ z  T! h! j! `. Q9 E! @
[ 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 G4 z+ S5 [+ U) T
sorry, did you post another question? your statement is still not clear.
) @4 |5 j m0 M
# U q% _+ t6 G7 s7 ?# e7 S1 n3 Ka, b, s, k are constants? c means c(x) ?
9 p9 l8 \! [" Q6 w9 c5 L0 M: t$ l9 z
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:
, f! n i/ I* @" [& @$ _0 L对不起，写错了，忘了变量c,应该是% v$ I; C& c# Z+ `
d [(a+bx) c] /dx = -kc +s
, B# f5 ]& N: q/ L) l0 b& O1 \# q/ g" K- j
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:
) e: T7 w1 ~- e/ K7 }! Ldo you mean it is:& Z. c, m. D3 d- O; K
2 r' h0 B' T5 F; p3 |3 s" d
d { (a+bx) C(x)}/dx = -k C(x) + s' @1 g5 B2 v$ y
$ E R O$ J7 r6 p$ V' D6 l
where a, b, s are constants, you want to know what is C(x) ?: ^" P- T7 T- B0 o, K. T

6 M% {! M0 @; T) l2 Q8 H3 dif 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 14:33:5 c3 [* N9 h! _
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:
$ |& O) A$ @8 X" @9楼的答案有点问题吧。8 }7 j, g5 b' @1 y
您说：
# I) N0 G$ N: z5 k$ ~4 M; Q1 L. }0 @- w3 P6 |8 s
d{(a+bx)*C(x)}/dx =-k C(x) + s+ W. r) F1 P/ Q9 e9 l5 ?+ ^7 l8 I \
so:
6 e' X' N! Q3 [1 qbC(x) + (a+bx) dC(x)/dx = -kC(x) +s) t' ^- U- K8 }4 B

. H* J0 g" w# E+ M) _4 \也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx* C* x0 z' K5 L0 v3 o! s
但这是不正确的
- M; G2 r. r; {. G3 g" j0 ad{(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 N8 E. B# E  C6 m, L....
! s, X2 ^( P  u% md{(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 20:30:7 L' R( F3 S4 r0 r% A1 q
这样算混淆了微分和导数的概念。
1 ~" e2 f) ^( q; p4 G+ ]: o0 Q% P! `' S3 w2 i$ q
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
/ ^; @5 S" N; H! q& t7 n( s& o$ ~( `0 h$ 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:
3 Q; w$ @6 W$ {* c& w* j" Y8 F( w9 m这样算混淆了微分和导数的概念。# A4 x4 v' X! S" K- _' w* _

) l9 ^# @( ~( Q2 l7 S- k: b; 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
3 a& N p1 W6 q* D+ J' k* [2 o/ @. j1 }6 r6 u
Ah, you made a mistake: it should be:
" B5 N0 C) S& z# x+ u3 G
: Y- s. U- `3 d; @6 td(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:; N9 @. c9 }: x' e6 Q: L
Yes, there should not be " ' " after "f" and "g", it is a typo. :D% T8 Q7 n( u8 `

! |$ H1 ~! P& W ]3 y0 qBut 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:
3 C8 |5 s$ i* K! U" c7 F- r# o
9 L2 w* h0 U3 \1 Q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 k4 |6 Y* j5 V8 c3 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:. H. A9 G% X' I3 h$ f/ T
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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