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:
% E6 Y7 O  n8 [5 X6 J' N请教大家一道微分题，多谢！* k+ O0 S, Q  T. b
+ C) f; e2 V1 n
d [(a+bx) ] /dx = -kc +s
6 h: ~: {  c* s' @9 R, bwhere: only x and c are unknown, others are all known,  requiire c = ?% J& q* C! K1 O( s

  z. v0 `( Q3 d. r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:
) R, [5 I* r* ]1 P0 e2 Jit is too easy:
5 f. p3 G! Z1 }5 s* z5 N4 {- H* M& L: l" }2 P6 Y
d(a+bx)/dx = b/ D3 s$ [/ Q/ V
" U6 Z2 C$ J: c+ l5 }
so* U9 ?$ K0 Q* M9 H5 L" j) J3 G

. o- B8 {( @: Nb=-kc+s
c, ^( x) q$ I5 p0 Y, ~$ u+ `; h
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 20:41:4 G$ a& O6 H& S$ Y' i
请教大家一道微分题，多谢！6 b- q% z: _6 v3 s

2 s% C3 _' Y( ]d [(a+bx) ] /dx = -kc +s ) z+ e& p& V+ ]* C
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
( i' M+ n5 g/ q( G+ j  T. x) Z7 [% H! k; B9 a& S8 y6 U
多谢了！3 H3 O- o9 B6 s- C

6 P, p3 }$ g' G" ~  W% c, G[ 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:, o# y+ v; e5 E5 q; B
sorry, did you post another question? your statement is still not clear.
! @" u: E. u% |; }6 j# _+ \( p7 o9 Q/ K$ x+ v( X
a, b, s, k are constants? c means c(x) ?
6 e# L. ?% D: l- o Q$ X2 H* o5 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:
+ V! {. c, p6 q5 f/ L6 C5 j对不起，写错了，忘了变量c,应该是) X0 p6 }# i" C: L
d [(a+bx) c] /dx = -kc +s
( h9 c$ `& F! K6 l8 W# o
5 _ U* i7 X" k! W: O: Z% q* D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:9 [2 ?& f- o. E: d/ H
do you mean it is:
; i" e8 n9 Y: n. t: W; ?/ {/ b3 u6 t: t* \ m
d { (a+bx) C(x)}/dx = -k C(x) + s
0 H, C# w% D/ w+ O4 e
4 W7 L6 n, j6 i+ F5 e) y0 wwhere a, b, s are constants, you want to know what is C(x) ?
e; |: }$ ?+ O- t7 p: B' V0 m" H+ J- N- J. q
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:
$ a, ~; u p% H7 ]% ^/ c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 19:34:2 H+ @! a/ Y) {4 D z e; o& m* r" I
9楼的答案有点问题吧。5 m9 H) K1 a) ^) a2 K( l$ e
您说：
) c6 H4 ~$ r! a0 \1 n& b
. B: M1 Z4 L9 v0 rd{(a+bx)*C(x)}/dx =-k C(x) + s$ v6 S6 o* c* O% X
so:8 |; T% X/ V! s3 t2 W% ~! P
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
1 v) H& e2 [4 D3 v3 R- H$ h; n* ]1 ^; `
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx$ m% W# E$ H7 e6 k" u0 M
但这是不正确的% \- J" f; l( N2 i
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
* I. e5 V- ~" |6 N& J; a....
1 y8 ]8 [* f8 x c& ld{(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:
( t, R( b0 N5 `7 z- b5 Q这样算混淆了微分和导数的概念。+ e- \* n i( W5 n \& x, A, ~
# g4 Y; i, k* G$ n+ v
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- y* @% p; w) e* h. n# n) R" u3 ]
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:+ }( \2 y' C$ M7 R) h2 \
这样算混淆了微分和导数的概念。, x1 e- Q/ [, |8 Z

% g9 } E+ U4 ]! k! {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: l& W9 K. d2 V t' X
' P; [5 G: {0 U$ o) [, S
Ah, you made a mistake: it should be:
) F+ O5 p; q1 R% R: D+ i2 o6 |1 u
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:
+ x9 L b* r: i% ~Yes, there should not be " ' " after "f" and "g", it is a typo. :D& ]& ]6 A$ y g" p

8 E- i& L; {, l/ uBut 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:
% d3 o) X" l0 j# K! g! Q* ~- j. ^' d# Y
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。
/ R/ F2 c' D' s7 f6 J7 z8 G8 k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 23:24:1 p5 I3 g: |, q1 H8 E
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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