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:$ Y$ I0 t" D( \% e& j
请教大家一道微分题，多谢！/ f$ @7 ^- y; p
! C& B6 i" }+ L% b
d [(a+bx) ] /dx = -kc +s
$ x; ~$ a8 s" i- @1 l$ Qwhere: only x and c are unknown, others are all known, requiire c = ?5 W1 Q& ~/ Q) w4 P

% O' P; m! f6 u+ L* i* b9 o, l! j. m! v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:0 X: @* P$ s7 N- D3 G4 A5 N- `
it is too easy:. `$ z1 s7 L D. {$ g- b7 \

) b; b. l7 M4 ^( r6 jd(a+bx)/dx = b3 b/ P$ @' l3 _; v
8 {% e1 m# g* T
so* \& T8 L8 P. y+ u9 h% D; G

' H) ?2 C; L' ^7 C2 _b=-kc+s/ m! m4 J0 [, a! n

! t5 h$ g0 R6 e# x( U6 l9 N) bfinally: 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:; S5 p2 L: k  n6 ^+ I
请教大家一道微分题，多谢！
3 v+ v" A8 v9 t+ t; f7 R! k
8 ^1 U% L" n+ [7 Od [(a+bx) ] /dx = -kc +s
7 I4 ~% d% h, S% D: x" swhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
% A4 Z, K2 o6 z5 P2 m3 k6 j6 ~- U3 L! w1 e) k; b
多谢了！5 [' v3 m$ e5 Y' F
+ O# k  t. O, x  N
[ 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:
9 h# I& p# H2 isorry, did you post another question? your statement is still not clear.0 ?$ \( Q. S; a: N* q

! o( y* p% j, G" ja, b, s, k are constants? c means c(x) ?1 |$ s% B4 {+ g. l' L0 A

" c3 y O# m1 J; Y! Dplease 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:
6 X9 G0 Z! h4 a+ y; u对不起，写错了，忘了变量c,应该是$ v; u8 ~+ n e2 ?& H8 E# A
d [(a+bx) c] /dx = -kc +s
7 O$ n0 i! a. `3 P: Y
/ j& X! d0 G! I' g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:
$ Z h( j; Z+ V# @' L* h2 ^7 Sdo you mean it is:! ?3 T# H: p( Q) F# A
# t/ m! K* l% P: F5 M
d { (a+bx) C(x)}/dx = -k C(x) + s
3 L4 O ^9 {6 I& b) D
- @5 F) ]9 O) \; C. Fwhere a, b, s are constants, you want to know what is C(x) ?
+ p" B! u) C5 \) U( b2 E
) T4 c. w) F3 Pif 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:/ x7 l8 p8 W5 Z' b4 Z Q
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:# T- X# q, ?# A0 L p. `$ h6 e- p
9楼的答案有点问题吧。+ f% z2 `# ?; C% T* @% d6 R, N1 w9 W
您说：
0 l, E; S( P: R- C# b2 B8 Y/ k5 _
8 r, ~! o, Z5 o+ Q4 d" T6 gd{(a+bx)*C(x)}/dx =-k C(x) + s+ t8 H5 D8 I' g# N/ g+ L+ H6 g
so:
. d+ O; ?' J9 d0 Y1 mbC(x) + (a+bx) dC(x)/dx = -kC(x) +s/ I/ t+ w& I$ i5 j }

) \, ~$ A) ]4 y$ j" `" o/ G- U也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
7 t# L4 r( B+ i9 m但这是不正确的
4 ]' h8 n- [: w, u* ]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' @" E7 A, }* L7 O g R
....
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 20:30:7 p2 T5 Q# q x' g# U: Y% A
这样算混淆了微分和导数的概念。9 E5 b2 b& _+ z' i# e
$ U! M/ l X* u4 ~4 [+ q( S+ 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1 ~- f# ~5 z7 R3 Z3 r2 Z
) k: O) T: y" f% }
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:. x+ T! ~4 @/ ?
这样算混淆了微分和导数的概念。
: I# p; L' R. d! K" H7 d
( X/ H4 B, S! E6 t; b/ f; A. h8 L. [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( o# H+ d! Z i& t* Q) [7 U. d& s

( a) b' ]+ a5 c9 `7 v v, }$ qAh, you made a mistake: it should be:+ N4 G2 k0 y3 z+ ~8 P( G. y2 s$ F
. K# Z9 v4 M; G# J" r( o9 K2 W
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:* C2 s; D6 a5 D1 K
Yes, there should not be " ' " after "f" and "g", it is a typo. :D- L- P/ z, B5 ^6 J. b% m3 m
8 e$ z6 X9 t9 [0 J2 w# L4 L1 y: T& Q
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:+ c$ C1 O$ R+ t; e1 U( ~! v

3 K9 T' V. Y$ t" H5 |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。7 w5 U& h5 }( X8 Q- I; [9 u/ D$ U9 ^
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:
8 |; R Z4 G) z9 L7 |' E6 r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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