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:
1 V7 o" x/ k1 M* [# h) {- @ j请教大家一道微分题，多谢！
1 @; y3 @9 G! ?, J6 c' {
& V( ^$ |$ c3 r9 w, Cd [(a+bx) ] /dx = -kc +s
' T* ^- ^' [2 O2 E" u0 h8 Bwhere: only x and c are unknown, others are all known, requiire c = ?1 V8 _; k( p' u5 V J! ]
+ J7 I( q. M8 }
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 E6 m$ m) \3 w
it is too easy:
. G. ?7 G/ L1 h" `5 B! g+ W6 {) B: L3 h$ ?; j0 A
d(a+bx)/dx = b& I# A+ V4 l6 \& q5 D

( L* Z3 U* k( Q t! C+ j: Iso3 b2 M/ U: d# Q2 F7 d8 g: C6 o

- D9 `: A- v( d/ t, @b=-kc+s
9 }( A+ I( E: P5 `* x2 D
: ?) e5 f1 }2 ^% o3 ~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:
1 t& [* R" j+ e请教大家一道微分题，多谢！" n+ y) F. @! K+ L4 I

* }% ~1 Q/ R  @d [(a+bx) ] /dx = -kc +s ) O0 m' F$ A6 F; c& X3 e
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
* N$ r! C# }3 Z1 Q9 a8 [+ |4 ]- R  t6 D" w+ O: l
多谢了！
- |, y/ R+ p/ Y3 k1 G
! H% |) x$ ~6 o4 Q' g3 S' {0 y4 y[ 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:
" I. [, N7 |9 y# Y1 M, H" e0 qsorry, did you post another question? your statement is still not clear.3 W0 t- w& w3 c1 r0 w! W& Q
! c* J T% u$ x, F& C
a, b, s, k are constants? c means c(x) ?; v' a$ d. m' e

3 R4 W& Z4 Z% E" D: i) Kplease 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:
% K- U4 \" ?* i- J9 n; W对不起，写错了，忘了变量c,应该是
4 m) M5 q; Q* B+ m P2 ~% kd [(a+bx) c] /dx = -kc +s 8 ~5 H* z3 F& r/ F- e$ y) m1 c

& H/ ^0 I) N6 u* ^& _. l$ Y0 F7 t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:- O9 g+ M3 W5 S1 k2 E8 c
do you mean it is:
; l/ V6 E+ M7 C$ w4 t6 ?; D' N+ E
: p4 i; Q- C: e+ x: G6 r$ cd { (a+bx) C(x)}/dx = -k C(x) + s0 v `' |4 P; l( G7 N4 n @

/ V- |% ]! c* s: hwhere a, b, s are constants, you want to know what is C(x) ?3 d$ s4 T) u, N! [
! i* `& V7 X+ I: M$ i; W7 ~( X( w
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:
/ j8 j: ^0 {) Y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:( l7 N7 V% U! P
9楼的答案有点问题吧。
" r) e) |; K) W# M6 }8 z您说：/ U% [0 H. G  T/ X1 `$ \! C
4 D7 G& {4 ?7 s; ]
d{(a+bx)*C(x)}/dx =-k C(x) + s# `3 r2 p: I) P; l
so:
7 {- C& d) Y! O" |8 c% d* nbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s& X0 s# o2 J8 Q, n. J* D

# ?- u4 e6 P9 @. e也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
" c  c3 w$ `5 q" t但这是不正确的
' c0 H! ^2 R( u. D" n+ H! U* Q! hd{(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 O7 s$ r2 D8 B....9 O2 Q" s i# T \- Q7 e
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:
; L, \3 }/ _! H( \$ b这样算混淆了微分和导数的概念。
, ~! L' @' s: R' v0 n- F6 O5 d& \) W9 b6 w: u4 q$ E7 u
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
/ ? M2 Z7 }; e( R0 b/ ?. p
9 r8 k) ?4 f- w. \3 A6 h- 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:
) D5 u7 h/ j1 z- s0 {5 q这样算混淆了微分和导数的概念。
: p0 g% \2 X: w3 l- r0 ~1 P; z4 d4 {# 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2 M& H/ v/ o% u% f0 v0 g
8 J; u; v1 {7 z3 A f4 i( @" T
Ah, you made a mistake: it should be:2 y; E9 }8 A5 X8 E' m+ d9 q
% f$ v2 d+ Z, N2 |( b( G
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:/ f2 s' }- u# q" y
Yes, there should not be " ' " after "f" and "g", it is a typo. :D' A1 `& P/ W& ]0 m$ C

. b7 b, P$ [1 C2 u1 N: |' n( E+ E& FBut 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:7 V2 `' N; a5 V3 R) ~7 u. b8 ~
- y* s) |8 v& ^
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。
( Z W) J) n5 Y6 x5 n! R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:( n/ P/ D( p* X' G9 G
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。

账号		自动登录	找回密码
密码			注册

数学高手请进，多谢！

浏览过的版块