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:
% \; P# `3 j. w  W  j* m请教大家一道微分题，多谢！) g1 l6 y+ X2 g( M' [" K

  b& O& H3 k+ e" y: c1 vd [(a+bx) ] /dx = -kc +s
# V& h% ^: W6 o, N7 _% nwhere: only x and c are unknown, others are all known,  requiire c = ?
# I, u) E, f: l( r: s) \
0 j; ]6 R0 z9 I8 @) c: W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:
8 c  i. o) w( m9 q. e& F: Hit is too easy:( W& f7 E0 Q% R" o0 H$ S; f/ H8 U

, l# q. {; E- y. }d(a+bx)/dx = b7 k" Q" F1 D6 v
7 c3 O" _" O' C- H+ _1 `
so
! G6 z. b' K0 X3 i, u, ?  E+ I& Z. a2 T, h1 @+ [
b=-kc+s
' Q! O5 O% w. b2 w
3 {' k4 h' T, @7 Ifinally:  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 f( N% b" Z. e3 H0 f
请教大家一道微分题，多谢！
, ?( N# G/ a$ s$ a" u
, N/ d$ _- C( w+ a/ w, td [(a+bx) ] /dx = -kc +s
4 T7 s. ~/ N* Y6 |" i' u- kwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
; O3 q2 D, a" v8 D8 k  L) @) i/ x. v
多谢了！. z# `: A( u- i  n8 z! g. ]

5 [' X% q. t& t[ 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:
* M8 N( f# e8 G8 h; N# Esorry, did you post another question? your statement is still not clear.# B8 g7 t3 I0 u! j

5 N. z3 Z3 ]* h7 Ja, b, s, k are constants? c means c(x) ?: S: D8 f4 s! u! |1 ]: v5 y( q
: Q, t. {- P( A5 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:
" t5 S! W9 F$ [' f+ t& x6 w对不起，写错了，忘了变量c,应该是9 v9 ]6 G# a$ r4 k
d [(a+bx) c] /dx = -kc +s
: U$ W. Q- @0 N( d% s' j7 |* i& {6 r0 A+ A
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:7 @3 Y: ~1 }, Y* q2 d9 N  ^3 b
do you mean it is:
: }+ _" N6 o* w. {
, i% u8 t) f- e+ W$ [d { (a+bx) C(x)}/dx = -k C(x) + s
5 I  V2 w+ Y  l! x/ d# s; U# H  C: \0 Z1 N0 b9 d: F+ `. u1 \! ]
where a, b, s are constants, you want to know what is C(x) ?
0 ^' C; i  D. H, c; L+ v% E% X, C; u* Z8 ?2 I
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:
1 Y3 v! R. \9 \: U) [4 `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:
1 f1 S, \8 I1 b% T1 Z0 @4 M9楼的答案有点问题吧。0 R6 [4 ^- g) d9 h' s% Q
您说：+ f5 L; L4 O0 I" \0 E" W

9 E- m9 X/ F& ?( _1 G, md{(a+bx)*C(x)}/dx =-k C(x) + s
+ o5 |* W s- s& {: q7 W: l) Rso:
# N' ~0 c* A* z% hbC(x) + (a+bx) dC(x)/dx = -kC(x) +s& Z, D: E" W; d' u: W

3 _4 }0 t1 w' _; M' L- `6 x# k0 U) [也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx( ^4 Y6 |# _; c
但这是不正确的! }" ^2 @, `+ u3 m5 _& @0 v$ s( ~
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& j$ ~! G, z; G& b6 x! Q* }
....
) B% }8 J4 X1 ~ W. L1 b3 dd{(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:
) s* o8 u' J: W, b0 ]这样算混淆了微分和导数的概念。
( R" E6 a9 Y4 Z' H6 V2 }9 ?+ ]# z( m- _/ f% v1 }
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
/ f; r9 H% E! K5 D+ ]9 F" p/ K
0 P" x+ Q9 q" \$ h- l& W* tIf 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:
2 J" b1 E. U1 }# @& p2 C$ l这样算混淆了微分和导数的概念。
& d# m: a `2 y8 w
. o+ f4 B& `. Q) ~: S3 p1 fd(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 i8 [( W$ {" b" s9 o! q

. W( A+ G, S3 J; m- }" \; VAh, you made a mistake: it should be:

( A$ R2 |5 I1 Y9 P; w" Rd(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:
, t4 x7 K, g9 {; RYes, there should not be " ' " after "f" and "g", it is a typo. :D4 d& v# C" b' M- b8 W) p0 N. o) m
8 k; b9 @+ k* [7 H0 n# K2 F
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:7 {+ _6 Q6 b7 V2 A" J$ b6 ~
, y% x/ T8 d) Y7 }7 \- ]7 S( }4 M0 M
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。8 `/ T/ _- i, }) q7 Z& m
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:
& q4 W) c( ^) [6 ~8 F3 {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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