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:. d4 A1 f, r& x) j# p2 W+ K) h4 i
请教大家一道微分题，多谢！
5 ^* a3 S& V. \6 w' ]
/ z$ W3 m$ ?* z. [6 _- ?d [(a+bx) ] /dx = -kc +s & |2 m& l% @% g) y/ m
where: only x and c are unknown, others are all known, requiire c = ?
+ q5 \) b$ v3 ?9 T2 \3 H1 k) _" q; 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:3 G5 Y/ C$ `7 i9 _/ n z. g9 M
it is too easy:3 r$ ^4 z% G7 T2 `; P5 H

1 q8 k6 V* o. M. K7 {d(a+bx)/dx = b7 J# a8 u4 Z" Q6 |3 @( y- Q0 s

; O2 p4 Z: a: H, b1 `* z+ l6 Gso9 }- A9 L- P4 {" m. M
5 O; A) i+ h0 o/ ~3 g, B
b=-kc+s
$ s- h# R! \+ l/ }( [$ B
- b I9 ^( g6 R1 F9 q1 gfinally: 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:* A- o( ^, E% ^) Z* M' i  B
请教大家一道微分题，多谢！  W" d( n0 N# z9 E7 _; A8 o8 ^
, m$ U3 r( V2 m0 o
d [(a+bx) ] /dx = -kc +s
. u, [! J' L  jwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?6 G( c! u. ]" M. h. }

' k; N# V5 T6 N( w) D1 V9 O0 g多谢了！) X; {2 u$ a% d% \

[ 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:" e/ I& \" W- ]+ j
sorry, did you post another question? your statement is still not clear.
# h1 Q$ R) d* j6 H! M k) L _ {) B& j6 ~$ g, [
a, b, s, k are constants? c means c(x) ?- P; m& Z2 W; E- ~, M& Q% w
, \, {( J6 p# K! i# R
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:
3 K6 J5 H% Q* F( Q+ [对不起，写错了，忘了变量c,应该是
* A: G# K/ B( \( f6 d$ H+ Wd [(a+bx) c] /dx = -kc +s
: f& B3 p i- N+ Y: t+ X4 f
, G! R4 z% Z* I" _4 _: y+ M7 _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:" Z0 X7 T; |( U" G$ n Z. w9 i! H6 j
do you mean it is:; r. |' r- t D+ K' o9 s# T3 f8 B

! O* v1 w6 V" u7 m9 F7 B9 id { (a+bx) C(x)}/dx = -k C(x) + s
$ G5 S7 [1 m) w! ?; @
1 G9 `" {2 j4 k" a; [! p( owhere a, b, s are constants, you want to know what is C(x) ?4 L, ?. q8 F& N2 F

% |" Q- {; A1 K, C( t! L& Tif 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:
$ m- N- ^4 g+ T1 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:
' c5 Q8 P. I/ d0 f: R9楼的答案有点问题吧。0 i' N% E  k: |4 a5 [
您说：* x) y7 g; Y( l2 B) i& N$ A/ T

+ v# n. C: c4 i# rd{(a+bx)*C(x)}/dx =-k C(x) + s
4 E! ^/ b; I, @! e6 p; m& D) {so:' ~. O/ y2 X9 L7 k. X$ _
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s' M; j6 ~+ K# h3 k+ N$ j, D4 e6 \/ C" F
( ], e# {# Z) W+ A! m4 W
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
: Z6 u! H0 G4 N$ T4 U但这是不正确的8 n# A% Z5 H  e0 g: x. u( o
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" ^. L2 ~! p$ S. N: U7 H+ z+ i' B
....
7 z8 k6 W- h: W* b& p/ m! ]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:
+ m0 s2 b7 Z3 s) g这样算混淆了微分和导数的概念。
; x4 T( o3 F! S% i: {$ r
- i4 T8 o( f2 ?7 F7 xd(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 O% k+ D4 B- _( [$ E- V) Z7 b9 \4 f w$ z6 g
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:
这样算混淆了微分和导数的概念。
* L$ E6 W @* i9 d& ? H, B% q% f. H
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
3 \1 b$ v/ z8 d
0 w% h) v* k( B% T8 N: vAh, you made a mistake: it should be: |/ e3 r- Z0 }9 ?/ P- D2 @

6 h& b3 F2 D5 Q- y1 x+ n# ]/ Y# Nd(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:/ B4 G' f0 M6 K' L7 y$ e
Yes, there should not be " ' " after "f" and "g", it is a typo. :D' s/ h X6 {! F _. U) @
( ?4 l3 ]" Z$ A9 s
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:
7 X' D h: C$ a1 y- P9 D; I
( Z8 G2 S0 F8 q6 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。 x" i+ I5 A" n& ^# S) O
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:
k6 ?& M8 U; x' j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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