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:! f* ~7 u8 f7 B4 c: i2 ?1 l @
请教大家一道微分题，多谢！$ n$ b( U: d; ]2 d, E+ r& _0 s4 h
0 O7 P3 b: A+ Q3 ^
d [(a+bx) ] /dx = -kc +s / O9 B) G5 l# X8 q, {: s" m5 {1 q
where: only x and c are unknown, others are all known, requiire c = ?$ B* ^' a8 j, ~! B5 v* t6 e! d

$ R" A( g! G( ~* ?% |" n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:
# }. {& o& c  C* t; X1 g" J/ ?& bit is too easy:0 h! [4 j! ]- W
! W9 M  V4 D5 a) T8 H4 i
d(a+bx)/dx = b
7 K. X3 d4 M) k* l& i. e( m3 A) x2 H
  O/ o; O' e1 Kso& f3 r4 R# u& c3 _4 F: _

* M3 |7 H. ]4 ^3 z6 o; T2 _: ~6 I& bb=-kc+s
2 s# t1 ~; V9 y$ M
; k, V: `* l. \% x8 `' \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:
8 f) j4 [. b2 ?" ~- D: u. Z请教大家一道微分题，多谢！# `) Q. |7 d! T+ V' W, P% {# b" c
3 l2 ~3 k9 E% E f e. j; x
d [(a+bx) ] /dx = -kc +s 2 _- k% J% f; b+ B" a
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
" D+ `) `6 g. C2 J" Z+ A! A9 l
( f2 a; Z6 r- c) h+ }: i多谢了！
4 W) D; @/ K4 j% }3 l8 M% U" [$ T' I' b
[ 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:  L- l3 ]* l4 y+ S& p0 l7 N/ v
sorry, did you post another question? your statement is still not clear.8 v  i" W3 R) f: r, |5 T
/ Z6 F$ Y8 }  o( h+ Q2 L
a, b, s, k are  constants? c means c(x) ?  f& t5 I* ^: ?
) \2 Z) r" Q$ H( F$ P
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:
! G/ ?5 D& m& i+ j5 }" F对不起，写错了，忘了变量c,应该是) j# E# \* e$ \) F Z
d [(a+bx) c] /dx = -kc +s
% h- r( h$ A6 m, c6 i4 j! u. u. _/ D, D8 J" O4 Y6 n
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:1 p! y9 W. S6 i' B$ s# H0 z
do you mean it is:

& a1 E; J' U+ s# g, a5 k3 b2 ud { (a+bx) C(x)}/dx = -k C(x) + s: H6 B( {! U$ z( {$ `

8 W- ?3 L6 M5 Q4 o$ r: f, L4 Jwhere a, b, s are constants, you want to know what is C(x) ?- q3 o* m i" m( E# E$ z
$ w5 i5 J5 ^7 G0 C+ m
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:
3 I/ `5 [. \* m5 E8 K" J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:- v5 p- f1 K* C$ \6 e( S
9楼的答案有点问题吧。: L e8 R6 j( d) N, s) j7 d9 a* V
您说：
6 Z+ \: k. l) ?: E7 ^6 @0 k5 ?' b; |
d{(a+bx)*C(x)}/dx =-k C(x) + s# t' c5 @' }. [; r6 n' W- X
so:
9 O" a( o7 \$ e4 ]2 w& A3 bbC(x) + (a+bx) dC(x)/dx = -kC(x) +s- h8 g4 S2 [% h/ p c7 f

2 h( b" e* U' R7 V" ^( `$ f, W- ?" ]6 m也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx7 K! f% }1 F7 a3 F8 j+ B# ~
但这是不正确的- p5 d( F4 K% Y% e" q+ m
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& ~/ u# f/ M9 z4 P; G/ ?% D
....
! X7 W* p. `) M$ C! qd{(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- \+ [' z" q2 p1 L& l" Y) {这样算混淆了微分和导数的概念。3 A7 K0 B0 Y& |2 s+ e
4 K1 {6 C- L& w* t" 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+ M) {- O- C9 P/ w7 D1 w

9 C. ?# c2 Y' r# \" X$ q: B* o1 J9 mIf 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:4 O5 o3 H; J0 Q
这样算混淆了微分和导数的概念。% {$ F- s/ ]1 h  ^/ Y2 u1 q+ M( `
5 b7 R( h; Q" Z$ K1 D4 n- ^7 T' N
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 s. {  @; z6 W/ t; s
# I5 \, j! x" `% x2 @( d, l' l% rAh, you made a mistake: it should be:+ z* k: ?* o3 M3 O; v, M3 u! j. Q

  D; V, d+ {6 o* m0 \1 l0 ]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:. s" L) C, n( L3 {0 ]5 C- I2 B
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
; i$ W- j% V6 C3 e7 Q8 |
4 X8 {: t w0 j( G, W8 ?* wBut 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:! E4 F' B' V" g! U
' s, L5 N1 Z# O0 Q# {
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。1 l ^3 x4 u. ?2 r- A7 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:
2 H: u9 B5 J& E. X2 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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