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:
. o7 a6 a" |* b: D请教大家一道微分题，多谢！  `& D- R+ L; o6 r$ [( K* Z

/ H: j; ~; C2 h0 hd [(a+bx) ] /dx = -kc +s ; S1 |! v5 C" A2 e
where: only x and c are unknown, others are all known,  requiire c = ?
) ~+ G' k. T6 v1 B4 |1 V* W" R) g/ k, }, f  S6 ^$ a
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:+ x6 R1 g' F! q  w6 n% r; A
it is too easy:7 l, x! t8 P- @
' j8 {" ~3 h  N' y4 N% [" y
d(a+bx)/dx = b
4 ]6 g/ l7 e" U" n  V! G( P
) O/ r( j, h" p% J3 H; G! [. ?( ~4 tso
% {4 n/ }4 w: ?. w# @3 y7 W
+ t5 ]! g4 I+ N0 U7 Xb=-kc+s. C! Z: A: X% t5 K( I
* s. f6 J- P8 }2 Q1 z; f
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:( w. J. v9 V  U; f  Y# M1 x* J
请教大家一道微分题，多谢！
& H, r+ N7 _1 K- L# z' F5 x- g  w# q5 q' @! D1 f6 N
d [(a+bx) ] /dx = -kc +s
6 {* z- R( _3 `9 m' Y3 Fwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?0 f% z% C5 F7 V

多谢了！
9 {, s* E& S$ N' X
' j( z7 Z; [0 a8 ?6 Q: U[ 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:$ h: a2 [% l# t# e7 \5 R$ }
sorry, did you post another question? your statement is still not clear.
1 Z$ b! H0 g3 w8 X4 p7 n, c% l: Q0 x$ _
a, b, s, k are constants? c means c(x) ?: `2 T1 {+ k& l3 E, z+ l
- @; `1 P, Q# b4 S. s6 `
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:* K" x8 t* b1 \4 k1 V- r1 [ o
对不起，写错了，忘了变量c,应该是* T" k2 t# k8 l7 Q9 J) a4 A/ e d
d [(a+bx) c] /dx = -kc +s
1 ~6 G5 {, [/ o/ S; a
) ^; ?1 l& j1 k. P2 {6 `( U7 Q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:
" Q* G( ?) U6 z8 |- Kdo you mean it is:
8 \4 e2 U2 y, u* H) [' v( J# Q" t9 f3 a6 M7 Q) s) o
d { (a+bx) C(x)}/dx = -k C(x) + s8 ^0 \ `! z3 c1 a

& C; X" p, E S" n2 Rwhere a, b, s are constants, you want to know what is C(x) ?
: q5 y& @& J2 B
4 B: v5 }6 Z4 x- fif 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 14:33:4 `4 ?% A# h0 o+ E8 K. d
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:
6 x. v6 ], z  `+ o9楼的答案有点问题吧。
, \. K7 y) ]# C4 x# L您说：
7 C" ]0 B/ B. {) `' p+ ?
  o! k! c8 E; L/ td{(a+bx)*C(x)}/dx =-k C(x) + s2 a. C8 l, Y3 l8 @- r9 h
so:( @0 m1 ^& R' h# R9 ^# I
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
- J! b/ M6 G6 E6 J5 c+ d* A6 J4 f9 a; h0 O/ a. a
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
, M+ t: ?( ]4 H  v2 S5 T3 f但这是不正确的" O! H' I$ u0 @/ |2 ?( \
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! Z0 _+ v# A$ a0 ]( D
....' A& S7 \; O1 d8 X: O8 x. l
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 20:30:5 ?0 r( x C6 g& w5 Z9 ?
这样算混淆了微分和导数的概念。
! P' b4 y/ u3 W8 D2 c+ y7 [1 U5 C' ]8 s9 B6 e# c- z/ w
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6 I8 s5 d( L( D. c& j$ c# Y
3 |+ z4 s! D* B% E# t
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:
+ O+ ?" j% D* ^7 `+ \这样算混淆了微分和导数的概念。* [: M; ]" ?, a1 g  l' n, i

8 z: G* ~- u8 D4 X% B+ t$ r6 Cd(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1 O2 M4 k) u3 F: o  X! Q5 l
0 \5 J& q. A: f* X! N9 C
Ah, you made a mistake: it should be:! J2 ~7 c* t. H

( E9 `  F5 P1 \' t$ gd(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:
+ A3 e" o( p0 n( E" DYes, there should not be " ' " after "f" and "g", it is a typo. :D- S6 e. w2 e& Z

5 n v( h2 q9 iBut 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:
: l& Z( O) r8 m' |; `& x5 L) a) x) o5 e, a
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。. B) N5 e, A! M. C$ x* \
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:$ s. Q; I% P4 P( I! q
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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