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:
2 S+ T* V* @$ u  K5 f+ a请教大家一道微分题，多谢！1 b3 U! E$ o/ }* T, Z7 Q# w

$ }, w. h4 Y- ?/ Hd [(a+bx) ] /dx = -kc +s   C# c5 v8 \2 |7 k8 ^) L6 E* _
where: only x and c are unknown, others are all known,  requiire c = ?
- H0 w& r% a/ u* V" q6 w" a. Q. J; R' s( w8 w4 w. |1 j
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:/ @8 d* c( k( ^
it is too easy:" d% T, U9 n5 {4 D

/ U% l7 h/ a# V0 u" V2 L7 N5 ad(a+bx)/dx = b7 f  f' s0 V* Z1 m( y
% ~/ [5 X; O+ i, w* |* ?  z/ F' J
so$ f. c* K5 @+ w) O  r( m
* \; e0 z; V  m$ s
b=-kc+s

% g7 b3 E1 q- V% `: b) pfinally:  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:' f8 @/ }* j( l6 D: N: R
请教大家一道微分题，多谢！7 _6 j4 q3 q. @- L
5 g+ g5 u7 I1 B0 V2 @) K, d% o
d [(a+bx) ] /dx = -kc +s 3 K5 k% Q8 q/ j0 w4 u
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
& s' ~/ I8 w8 z5 H" {" R. D5 B0 k0 n5 o; w
多谢了！
! D# N. @7 O) P0 G( z/ J
) u# `3 B( X" x1 N P# ]* @+ a[ 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:
7 [; u% q# i( n( @sorry, did you post another question? your statement is still not clear.
3 {( T# G# l, g' W: c. N$ v1 f. ~- {2 I' g2 y' a0 [: w% r
a, b, s, k are constants? c means c(x) ?2 n+ m2 v0 ~1 _! s! z1 [
8 ]$ n# F, J4 K9 K. h& T
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:* P# k4 o4 I7 H  o: ]4 D
对不起，写错了，忘了变量c,应该是
$ N* y! D( J  f4 s; d" @9 ]/ Vd [(a+bx) c] /dx = -kc +s
8 C$ y& M, f# t6 O  l( u' u$ I, w$ [2 t1 N# }
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:
# T  q$ P# H! J1 l0 J9 Xdo you mean it is:; O. v* j! D9 L5 @& J: U; E
* {5 k  I( X% y
d { (a+bx) C(x)}/dx = -k C(x) + s! U( R5 R# I! H8 B; L4 g
/ _+ [8 V- w& V; m
where a, b, s are constants, you want to know what is C(x) ?
3 D3 [2 I  Q! F
0 n0 h; v  _  U% I' {& qif 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:( C H' I2 ^9 B% W9 k# L+ A
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:
6 z3 s; R/ ?6 X: w( }( p3 }/ e8 j) `9楼的答案有点问题吧。
" d( P( G6 i2 D' H3 j6 }# r. b& Y您说：4 _0 b* k- y6 G. x+ n, ^/ n

5 f+ V! a) W; i# hd{(a+bx)*C(x)}/dx =-k C(x) + s; \# c1 ?1 d2 X" Z. K' U
so:
2 ]0 m0 M, Y( J$ WbC(x) + (a+bx) dC(x)/dx = -kC(x) +s- N/ F+ u: l: {$ d3 n+ d/ V( E5 i
: E0 V. n8 j1 s3 @& V
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx5 g$ J0 h: i! t" j
但这是不正确的$ P9 X% P8 J; Y# x
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
# x6 U% @ s. l; E; J, x....3 ]+ u+ `0 n( N5 c& B+ |$ }
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 G D& h$ K; D2 ], d- o这样算混淆了微分和导数的概念。/ k$ Y0 f' y+ j
1 i8 p# S8 ^) l( ]+ h- d4 t z
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6 ]. C% a5 d0 q. w
! c% N) J0 r% e, p# m2 O, Q
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:. E3 j2 H. B9 V5 t9 M
这样算混淆了微分和导数的概念。
9 D% a8 }" L: `% T2 R3 Z7 M7 l- }( l2 d' P9 O
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, ?* @" H; O0 L+ K+ K
+ M' U, C9 S* F; W
Ah, you made a mistake: it should be:
?# ?. _: R0 b8 @; j# {" I3 _# |! D; L4 T2 P6 f
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:
9 y( w5 ?5 j) ?( p2 t4 O) @2 dYes, there should not be " ' " after "f" and "g", it is a typo. :D
2 F- z0 h/ i/ K0 p& z T0 G; x/ y
0 ^! D' q% v) @' |/ P4 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 ~/ v* X& j( P' q" }2 i. G
+ q3 e A: `! |0 u1 I/ f! e: X, Q# w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: L; q. q# r0 ~9 X3 p& `0 `- 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 23:24:3 C B! O. G/ A1 j/ \2 h7 T3 M
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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