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 20:41:5 |1 u# Y5 u0 o6 {
请教大家一道微分题，多谢！
% V0 d: i; [' G4 `, S9 y% ^8 W# Z5 a: U3 L; T" Z# U( D
d [(a+bx) ] /dx = -kc +s 3 ]; d W* E O; L* Z; A4 `
where: only x and c are unknown, others are all known, requiire c = ?
. b2 O; A1 U) R6 }, J+ j$ F. I: P! J& k% F4 H
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:
+ @+ a5 q1 j+ {: p1 Bit is too easy:
$ d3 o0 l7 M* H: s3 t
4 l0 Y. O' y! K' X  ~d(a+bx)/dx = b
! |. [/ G1 F7 Y$ ^* `6 s1 M  b- d! g4 N9 z
so/ i3 z- z( D- W7 o  X5 i3 o  [
8 S8 F- H- ^/ E
b=-kc+s
/ G4 ]+ {! x+ f8 R! {9 V# @$ h# P; W4 }+ O) W: q4 Q/ t3 D
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:+ I9 F6 I) d) x* G4 P  b3 c# ?. {0 Q
请教大家一道微分题，多谢！
$ M/ x, {6 ~! _0 n* U8 d1 o6 i) Q: o. ]) i2 O
d [(a+bx) ] /dx = -kc +s 6 f) e+ W7 L0 ~' \
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?0 u" c6 O! ]: r; }7 l! S+ E
' Q6 x9 G- L. C9 T5 D+ U! S+ \
多谢了！
" T& j) n) I, K  }3 h' N
" \! \. U- d1 \[ 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:
) f2 |, x) O: ]3 n$ p4 D. Ssorry, did you post another question? your statement is still not clear.' E: I( b! `9 t8 G
3 [0 N# {& v7 h I9 s
a, b, s, k are constants? c means c(x) ?
: H. g( Q: d/ J" m8 S
. p4 g/ a/ Z. ^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 12:30:4 \* _: a/ {. A* f; k x
对不起，写错了，忘了变量c,应该是8 h# I/ i( G* v- [% a" N
d [(a+bx) c] /dx = -kc +s
, J7 M0 }' k# ?
# G7 f7 T4 P/ P' X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:& M7 D% ~, x$ P' b
do you mean it is:
2 G6 ^  Q5 k# Z' W, r5 k
1 U" d8 {# M$ i7 Z! \6 _d { (a+bx) C(x)}/dx = -k C(x) + s
" [( F# ^0 R! u5 \% I8 b
8 V  I3 V3 [* o* e1 xwhere a, b, s are constants, you want to know what is C(x) ?, Y8 ?# d$ w  p0 p
. V7 N( w- j  _, L5 ?
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:
+ D9 ~! H0 i3 O) j* ?8 t3 E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:
# s7 K8 x4 T( U% R5 z9楼的答案有点问题吧。# [3 O2 M: y: {9 w: d+ D
您说：' P5 Z2 V( @! m1 a% e
/ R; \+ Y) u6 r) Z' l
d{(a+bx)*C(x)}/dx =-k C(x) + s
; a8 b5 s* ~5 s8 N; Yso:0 ^7 |% ^; n1 ^5 R3 w3 C, P) a
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s8 U* G8 j# I: e( a
9 Q+ n. J/ ]: Q* f8 U$ R
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx  }; G; e8 k2 k; F
但这是不正确的  C; b" l9 k, l% P& W
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
n& E# o* p- x; }8 s" L...." G( v6 q8 b1 j- K$ n/ y7 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:/ T3 K+ m+ c  t
这样算混淆了微分和导数的概念。
' h+ t4 R! j/ C9 D8 v$ f
$ B8 f5 k/ W2 T# V& o, h6 G6 Ad(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 C7 c9 O# B; m' }  d% u
8 p+ g/ S0 ^  v% x5 sIf 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:: S$ u* u( g/ B3 Z7 h+ o9 C+ _
这样算混淆了微分和导数的概念。8 D- W' L! e$ V- H2 [% n" f! Q- K! w/ s

F4 e' D6 e- d N. |! k* Ld(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
4 K1 f8 D; Y% s! N6 y8 D: u/ `9 t% p# w+ \4 Z
Ah, you made a mistake: it should be:) ?4 _! {( `1 {1 m' j. L2 i

5 c( Q' a2 J- K# ~. Zd(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:
$ y3 c! q: m! Y+ hYes, there should not be " ' " after "f" and "g", it is a typo. :D; `: x$ W8 c% c6 q5 z$ b
( R d/ x3 d l) [6 v
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:; j; O- A% F5 v) X: ^

' L+ t* C7 w6 ^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。' t: |2 L3 Z" W2 U( J
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:
4 O5 W3 K" m* D; S# ?6 N, Z. ?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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