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:
, E" U% Z5 l7 r! e3 \7 f1 u# P请教大家一道微分题，多谢！
) g# ]0 T( F2 K$ m1 _2 E0 ^! @
1 d+ |7 P  ^) ^+ T9 h. R! q. R: @$ {d [(a+bx) ] /dx = -kc +s ( M. F9 _0 I$ b2 e
where: only x and c are unknown, others are all known,  requiire c = ?
% T  s$ Y+ N, U, f+ F% Y4 s, A; R; s) n' T- `
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:( c: C3 O, e. x4 i
it is too easy:
' ~7 Y, l, N+ o$ t+ Y. i' \" \" b# m/ }( i+ B7 T5 [% z
d(a+bx)/dx = b! N0 G* Y& U- ^% Y3 a% F5 Q

% D+ l: g1 c/ n5 z/ gso- I2 p2 ~$ _/ @- F
: X) }( Z( W0 t0 b
b=-kc+s
$ g9 c: J* H2 }. _; }7 z% u
# m6 z3 p) t$ a/ E* Y: u' mfinally: 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 20:41:2 n5 t4 }, p2 L3 D
请教大家一道微分题，多谢！
+ p6 c. c, g* H% K0 O4 h' r
+ ~8 k* `& J8 |d [(a+bx) ] /dx = -kc +s
( H1 [9 M* t2 b: Y# V8 ~where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?7 X% l5 G* U7 H3 k1 s
* C  O+ j* L& e4 Q
多谢了！
# C0 P* ^" x2 Y$ ?2 l* n: A2 F  p" k, F4 C
[ 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:
0 Y4 F" ?, B1 N  j& ]% H7 z& Qsorry, did you post another question? your statement is still not clear.- n1 P# Q; X  a6 E% x
2 u- `) p/ ?! [0 Q6 o# j, L/ c
a, b, s, k are  constants? c means c(x) ?
; h  O8 @; s& d4 l( I, g+ [) v+ G
+ J# n7 \* |9 D! r0 I, N# i" d& y! w" Kplease 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:
8 N) O+ }8 @, S% o对不起，写错了，忘了变量c,应该是
' h$ z( `+ y( Cd [(a+bx) c] /dx = -kc +s * h V; i, K7 f1 r0 A6 l) L

7 @! Y9 F, D3 V( c7 O2 J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:
' f# s! H3 A9 p7 f/ g+ q0 wdo you mean it is: l. a; I7 Y: g+ E3 R0 r
6 w3 [; c; [. W
d { (a+bx) C(x)}/dx = -k C(x) + s
) V: _# u7 x4 f3 M7 T
% m9 X# J4 ~2 B8 L) f7 qwhere a, b, s are constants, you want to know what is C(x) ?
" x! S1 G& s( o8 g0 [2 B4 A
% c# M+ ~" ^% D/ V0 w' b& ~3 |- _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:% | i. B) q3 d9 |
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:
2 l" U% o7 @; {  D7 U9楼的答案有点问题吧。* v" k- E) H5 {" K
您说：
* Y1 H: [# c5 N* ~
6 ]: l% s5 o+ z' w- m% p3 Nd{(a+bx)*C(x)}/dx =-k C(x) + s: A, [" d+ @0 s/ |$ ~* f
so:( ]# N$ I, l! S* D! w
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
6 t) R& z! u8 v0 n
9 A: q1 G5 P2 P. I也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx, x$ j  A7 Z' x3 v, e4 ^) B
但这是不正确的
4 k- e4 b4 n8 t* P+ R! A! }: u5 w# Y3 Gd{(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
8 t1 O* [* s# v....
; y- g# ~3 v* l) h) _1 w7 |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:
. W& t: a5 N* K) Y, S9 D这样算混淆了微分和导数的概念。
+ |) z( I) b9 g# p0 T% w- j# Y( q# V3 C) ~+ T+ i p
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
9 A9 s t0 X. z( \; @% @7 o* F/ ?; t1 q2 q0 u2 b: t+ w5 S
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:
' i/ ]' L* s" K  K这样算混淆了微分和导数的概念。
- E/ F8 i3 D  a0 f  ^: S# e: j9 E; x: s) v+ G. 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
  U- ~6 u% S! c+ ~8 v  N0 f/ X: ]/ {. ~# V
Ah, you made a mistake: it should be:, q" f. O  Q6 J; D. x* ]( n

& A! S9 j' I% \, T& dd(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:: |3 E/ N: g/ P% w
Yes, there should not be " ' " after "f" and "g", it is a typo. :D6 u+ e" E: V# U. H1 l P! z

: d0 g5 }' |& n. k0 p1 qBut 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:
7 u# n: A. Q2 I6 N7 z
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。 s# n% t9 g& r# K- J/ q1 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:- C8 c/ d6 x0 B$ l: y: \: C; U3 D
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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