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:* s4 h' Q# q( B/ G- [
请教大家一道微分题，多谢！
) f0 i* [0 F  Y" A- v
0 y9 ?" F3 B) qd [(a+bx) ] /dx = -kc +s , y; V0 I* @1 t4 k  B& Z5 U# [
where: only x and c are unknown, others are all known,  requiire c = ?* n, @6 @5 V! A) o4 s4 G1 v" z$ c

- i+ {6 }) {# u, c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:
" v# I5 N& z1 X3 |% B0 E8 Fit is too easy:
. z' w6 E3 ]7 T" j1 l  x1 N) w3 [' m- c( E' V! c+ N6 i
d(a+bx)/dx = b
& B! z- C1 t7 [* g. R; d5 m7 v
3 W  l3 @% b6 A4 t. D- vso/ ?! R5 q- x. k7 k7 P+ P2 W
6 W6 _- z- z* B: J; Y, e9 o
b=-kc+s
9 X% O7 X2 B5 o1 w% i; m! G
& G! c- {2 ]& yfinally:  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:
5 V6 d! S/ r2 c$ N请教大家一道微分题，多谢！6 U; K( W$ E+ ^; B

- _. s* c  p, _# d% T) \9 \5 dd [(a+bx) ] /dx = -kc +s
/ [/ P. V3 A- [9 ywhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?# Y- M' }" c5 l3 A6 a) L

2 [( y7 E# g, B多谢了！+ z) c0 ]  j$ A5 b

* \5 Y  R/ y$ u+ _/ @1 o: w[ 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:9 l' {, r* ~/ m: r5 h4 A
sorry, did you post another question? your statement is still not clear.
7 X" f' z. \. c3 \3 r S9 V; \- h5 \9 u% G! Q4 K t0 m s, n" _# `
a, b, s, k are constants? c means c(x) ?
0 v/ h+ v/ c* _- k0 J* K. F; D( j- B* H; E
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 i i4 h$ Y) r2 {% k* i
对不起，写错了，忘了变量c,应该是
: G% I: w# r/ O, d4 f) id [(a+bx) c] /dx = -kc +s
; ?" _+ R1 k" _2 a2 l: u* T
) N" z/ k9 c- x2 U1 `! F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:
0 b# \( @% a8 Q5 f5 ado you mean it is:
4 b/ h, B$ g+ B) b
: L c! V- A9 O2 B0 w kd { (a+bx) C(x)}/dx = -k C(x) + s
/ {6 O! B* f4 q5 n, w
. n4 _) a5 e3 }' @) Jwhere a, b, s are constants, you want to know what is C(x) ?
+ n/ ^. W6 ~9 _* i& t3 D3 s) Z' S7 U) 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:
v$ C: o9 b0 o _. S% W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:
* _/ b8 B; h8 X/ ?/ I9楼的答案有点问题吧。
3 o/ i  g- C* X6 ?  y2 k" G您说：
  P5 X% t  J2 P& w* e2 H2 l; w1 h' G2 h) v- o
d{(a+bx)*C(x)}/dx =-k C(x) + s
, f* j1 t  R3 o% V$ Y# ~so:
) Q! w! M- `# I9 x: fbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s3 \2 N6 r5 R2 N" g

* o( Y8 C& ]2 x# y# p也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx; F0 e& t* p- x0 T$ I
但这是不正确的$ `% j5 o0 A- ?
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
  f# v3 @9 M4 Y# X% L; v....
7 V  U* K" V0 [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 20:30:0 k) Y; g5 B# B1 T; `
这样算混淆了微分和导数的概念。8 N0 X/ g2 ]2 U
$ _: }2 r5 X+ \: G
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
# A: C& a. B) {( G
, F2 O p; s2 v! m0 O0 NIf 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:
7 D/ k+ {2 G h8 B这样算混淆了微分和导数的概念。0 r4 W( S- r. \" O

# h3 Q$ V0 e# b) Q5 rd(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
! I+ S5 C) F; U2 F6 `
+ d$ x3 S# {6 }% g9 \5 e. PAh, you made a mistake: it should be:* o# a5 H9 v! k4 a6 [$ k# V

. u6 ^0 k9 o+ [! e B) }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:
1 u- ^+ K# Y' C) V FYes, there should not be " ' " after "f" and "g", it is a typo. :D$ L2 S& b% z6 R+ m1 g* n
; }! N8 t J+ v2 A# o& |
But 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:; n, X/ V4 ^6 a! V/ i* i- V

" Y! C% z# L9 z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。
0 ~+ C% j: g# D% \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:: ]$ I/ u8 S) m! a! l8 o
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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