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:% S) O* R! p, h( C
请教大家一道微分题，多谢！) H/ ~0 i# F3 p1 k6 @) }( H
* z" D* L* E; a" d8 T8 t2 m, ^
d [(a+bx) ] /dx = -kc +s
+ d2 t+ P6 _4 A0 v5 ]- kwhere: only x and c are unknown, others are all known, requiire c = ?& l: P5 F4 T7 t( I7 Y! r& [
$ Q1 n; Q& l0 M4 |: D) W/ n. J, D
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:
  ]' g' ]/ V9 S9 Git is too easy:
0 i" D4 d4 N! U8 E5 e2 I6 R, q8 y8 ~& N/ v0 _9 K% R6 I+ ?
d(a+bx)/dx = b
3 d2 q. C+ v& ?" q8 F+ l( u9 x  z! ^+ _* g* N0 k" w% m; p
so
: p$ d5 H1 c' E. M$ \
; r8 o1 j( R" Ub=-kc+s
, g+ n& r& S" b6 z! T/ C' a# {; }4 x
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:
: j7 ]3 [6 z4 D, R7 ^请教大家一道微分题，多谢！ J% l6 O4 W; t- A. u5 s

5 j' H' {2 p& h- D+ _d [(a+bx) ] /dx = -kc +s % F5 s/ w& V) d: A: e" _
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?9 d! { t, Q5 N
0 `' L( q& k( {' F* Y
多谢了！- \* ]# f: f6 q
2 \! i7 `* I1 L
[ 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:
. x1 `" H3 X* O( f$ ~& a3 n, D5 K& msorry, did you post another question? your statement is still not clear.6 v" `9 g" }* F4 x1 n+ }

9 T; z! a8 x3 Z7 @ La, b, s, k are constants? c means c(x) ?
' j! I$ ^- U, u: u( z! n# k+ D! u3 ~6 R0 a
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 12:30:0 T% h% s+ ^' z" x6 _& r4 Y
对不起，写错了，忘了变量c,应该是7 F2 K0 c+ D3 z5 y6 J% h
d [(a+bx) c] /dx = -kc +s
$ D9 a K9 i. `4 X
( z# l0 _" K! N/ E$ c* ~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 o) Y2 p. s1 P3 T, P/ ndo you mean it is:
# [4 u1 _9 z+ }- F) H8 J0 A. I6 O" W$ }$ U: g) E
d { (a+bx) C(x)}/dx = -k C(x) + s
+ ?) j& b+ f( d3 }: f' _6 x3 ^1 R( C
9 }6 r7 F, q+ G$ e% E; Rwhere a, b, s are constants, you want to know what is C(x) ?" X- X7 N \+ b: `4 ~
9 O; m& P# v* E) R
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 14:33:9 y, t$ s( j6 U9 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:: O8 J* Z8 g" K( f
9楼的答案有点问题吧。
9 Y5 F0 j, w5 a$ n' W/ n您说：
, L, l  j: C5 L
1 u; {$ t9 A5 }# q( L  md{(a+bx)*C(x)}/dx =-k C(x) + s
7 ?% }- z! T- |$ c! o! ~so:* ?  H  A% d) J# f( F# C
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
) R+ S* Q" I- z2 V0 A0 R! K) J) j3 l: o3 e
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
: @4 B; K1 U6 S) {6 c3 y但这是不正确的8 o# v3 U( L% q# z' e
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9 I$ n+ D- g& S l- h: E/ p2 q1 q
....6 w8 v; k4 `+ `* u/ @+ @
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 20:30:4 z( T, D; Q  L% C, C
这样算混淆了微分和导数的概念。
* G2 B  q4 |4 }% q" _8 S
) E1 K1 M& o6 ]) W( [/ i1 n! vd(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
0 s" i  ^& T/ k. Y9 h) k! k+ z" ]$ T/ }! q/ W
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:
; b' \- j4 k B5 N5 v这样算混淆了微分和导数的概念。7 k! ]0 i$ P: ]
) ^! L) @1 _+ m8 ?; \
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
! N3 G, g2 k. F8 l7 ]# q, W5 s
9 o, C. B# {; {! ]( p; EAh, you made a mistake: it should be:
. e, O( d: o1 z: n2 `; P2 ]) H0 S5 S7 {7 m# h- c+ Q- R
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 B( Q# h! G* Q! I& }! aYes, there should not be " ' " after "f" and "g", it is a typo. :D+ y; N7 _% w m1 b6 J
; P& V1 F% k& `' Q- N5 H; d/ T
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:
* @- M5 V" J, Z1 ^/ n$ z9 r/ b" ~' |+ X
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。
- G W% `- Q5 h2 w v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 23:24:0 b S x- U! ?% `+ d9 c( k
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。

账号		自动登录	找回密码
密码			注册