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:
; J4 z( ^' h% ^0 X3 w请教大家一道微分题，多谢！
* r; Q S$ |$ D( r/ S6 {, E3 d
' ?8 {3 k8 q9 d3 b* _- B* ]d [(a+bx) ] /dx = -kc +s
! @: j* e" X: f3 \$ zwhere: only x and c are unknown, others are all known, requiire c = ?1 v! d( F- b1 z, U# {2 [ ^

9 @2 t2 |. W8 V" q$ X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: s3 |7 t" T; x: j( W
it is too easy:
4 Q0 M" ]4 h% W
* u4 Q4 r0 X( m  `: C. z' g% hd(a+bx)/dx = b
/ H9 M5 e3 \; T, R- R4 d  V, o* Y8 n6 R6 R! k. p) w6 X. I. b
so
0 o! n9 X* X; Y: s; _7 N2 n* N3 J1 k. F  e; S
b=-kc+s
" U9 ^3 I. E$ n# N; ]: ~
0 Z5 Q  ~+ L2 |3 q' Z7 B0 Kfinally:  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:
3 v; j3 x3 W0 Q; A) O请教大家一道微分题，多谢！
: Q# N- r1 Q0 a. J4 c( B
) O2 J7 b9 y9 G0 f; K6 Vd [(a+bx) ] /dx = -kc +s - G% u9 n7 h3 C3 o' `! C' ?
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
- P$ q" j0 B  R- @0 r8 M: ]0 M( Y
( X6 a$ _; d# a/ S2 S. {多谢了！
- O; m' N! J! N9 m  J
, k0 J& A1 f  H9 r2 q' o! n' 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:
! d; p! r" U, O9 P: Gsorry, did you post another question? your statement is still not clear.
! P) c& R2 V- i- f/ w4 m/ J5 @: A7 m
a, b, s, k are constants? c means c(x) ? R. E' ^0 l1 \
0 n: z# g4 G/ @9 ~/ T
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:& F3 L# b/ }7 Q7 g
对不起，写错了，忘了变量c,应该是
H! ], X- v0 ]& |d [(a+bx) c] /dx = -kc +s
! A N( h' f: b: O$ G- t# W, | [
4 |1 ~( a0 w. ^0 s$ p3 t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:/ w8 y0 P8 t. M. C
do you mean it is:: Z8 g. u6 K& Y1 ]: Y9 p. A
+ v" z: b% O% i- p u# E( f* `) D# U
d { (a+bx) C(x)}/dx = -k C(x) + s
5 t' P8 p8 z$ o+ K7 V3 ^. d+ x/ o P) A7 X& E( O) {
where a, b, s are constants, you want to know what is C(x) ?
: {/ W" h' X7 M7 ~+ `3 T
) F+ r$ T- ]( u0 r$ `4 ]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:/ ?% O) Q& H8 H3 \
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:, C; [8 S5 e0 Z5 _& v
9楼的答案有点问题吧。# l+ _5 P! x* L: g9 M
您说：
$ [8 `9 C1 q' l2 m, `3 C5 N( E0 W  O4 T6 J/ e, D
d{(a+bx)*C(x)}/dx =-k C(x) + s
: W! v. k% n2 d1 _so:
) h) d8 K: V# l- R  m1 L5 r- NbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
  `* C8 D0 U3 G! @0 X
6 |* Z' t: k! q( p4 d也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
0 U, Q4 p9 d9 |* h0 y但这是不正确的
) j( O+ n& L6 c! A& R  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
0 _" s/ G8 s! n, C....
+ w2 g. k9 r% {7 |8 ~9 p8 l% Nd{(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:
' f, z; } B p7 h) `2 `# ]这样算混淆了微分和导数的概念。+ f9 L/ ]5 Y0 b
1 k' }. `, y4 m& s
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. q3 N: ~+ J, Q0 D" X

[( B3 ~; v4 D! Z( C) EIf 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:
/ \, _) D1 C. a# y$ G这样算混淆了微分和导数的概念。
) _3 G( O; d1 u0 f9 H+ Q) |7 Y/ z# q$ o5 S* |0 C2 t5 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 z" d/ c( ~6 \" ]
8 H! T* A, N2 {* o6 E+ f, Q t& cAh, you made a mistake: it should be:& f' q0 n+ [7 ~# T! [! B

2 X# R+ S Q+ A6 v# @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:
& Z$ N' o& w( e3 [$ o# m. }) YYes, there should not be " ' " after "f" and "g", it is a typo. :D
\$ f, `" q9 o; R5 l/ w- D
0 `4 }: G6 R, w0 O1 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:
' M5 W& g7 \4 q( n
. P0 `: t) m: A' |% l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。$ P: e' Z' q' [& d* g
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 23:24:6 E( T% c6 j: |0 p8 @
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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