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:
1 Y) }6 z9 i5 S: A! j请教大家一道微分题，多谢！, T) K  k7 Y+ ^) @3 ]2 ~
+ |# C# K6 Z0 i8 g' [
d [(a+bx) ] /dx = -kc +s " s" D7 `. v  w9 \
where: only x and c are unknown, others are all known,  requiire c = ?
$ v" u5 B; B$ E$ F3 ?
; ~+ [' g# ^7 ~, z" p0 ?9 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:3 X; M5 |- u4 E3 L& g1 r3 ~
it is too easy:7 T0 l6 L2 B& \7 N& E) t

6 ^- | g [1 Z" H& d* @$ ~d(a+bx)/dx = b
4 h( y$ n+ x- C8 \4 W8 g4 o& p! L7 G
so
* x/ _+ ^/ d: M$ X$ h- F, F" t+ G
b=-kc+s
+ Q6 \5 X$ U3 O/ j: c8 K9 O8 S3 j3 `' o& M3 y: H8 T
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:' x6 k X- Y8 K: M
请教大家一道微分题，多谢！1 t$ E& R& W$ X4 X1 T- E4 s0 H
5 J- J* ^2 m# P9 N
d [(a+bx) ] /dx = -kc +s 5 y9 d" ~3 b7 h& r) e8 Z. C6 }
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?2 {5 G$ X: F" Z" \6 c! }. K7 p
% L3 q- a' l& K% I/ V
多谢了！" A, y2 B; P# X# m0 @. u- ^

# X7 X& L" ?# W[ 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:3 D7 {0 E; H* s' Q( h, |1 h/ J
sorry, did you post another question? your statement is still not clear.# Y* Z( H! @7 G% y) L' }4 S; U
& n$ ?: r! ]: E; A/ p- [
a, b, s, k are constants? c means c(x) ?/ m7 l9 Q; G! H6 M* M
- J. c6 ?1 n% K# n0 ?# ~ n
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:/ U N" V' Y' O5 v3 i/ g! q S
对不起，写错了，忘了变量c,应该是6 q F8 x# _; D3 x0 \" {, o: i7 V
d [(a+bx) c] /dx = -kc +s 3 U; u2 X' ^+ N$ ~
: I- }3 j* l7 y/ a% `) y
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:( h, L4 |2 L' L) t
do you mean it is:
! }- |2 [1 x4 p0 U- G2 D. Q% s5 `
c* z6 P- K1 p" ?( M7 c" |d { (a+bx) C(x)}/dx = -k C(x) + s" j; }$ B2 U1 g2 A1 I% g# c
8 E. }) m- ?- u* [5 _4 J& a
where a, b, s are constants, you want to know what is C(x) ?6 g! Y# U9 U2 L6 F! c; K

( Y# U W! Q L' @, R; A/ r) Lif 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:; Y# Q3 R: W% k
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:
7 @; H. m( U4 t) K9楼的答案有点问题吧。4 [+ C6 Q2 X N7 [3 V' b
您说：
' _; Y' a2 N- K4 f6 W4 O. i* ?" W- I; G
d{(a+bx)*C(x)}/dx =-k C(x) + s
4 Q) q1 ~: E& z/ N" v+ Kso:
& p+ G b$ `' h# R# G; W. ?bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
( `) H6 h, x# n: T/ M! V$ M' v, J" U' {' E
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
' |6 l2 X. f' r9 F7 O6 Z9 U, _但这是不正确的
; Q9 }: S, L3 M- ?" w7 x" Z3 Xd{(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% g3 z0 U( D- }( a V$ ?9 B
....
, s; E5 J% {* d g# l% w: N0 wd{(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:) E" }( H; F4 e; Q
这样算混淆了微分和导数的概念。
( ^- t5 r- a) W, B4 }7 C6 w6 q7 v2 q/ F y& e* o6 K
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" s: ]* k% z5 q/ H: p
9 f" _" e+ L5 ]1 @9 y/ t, m) 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:
7 Z- z- P9 K% ]; j( I这样算混淆了微分和导数的概念。
" B8 N" {! M9 b, c9 B
9 Z0 o+ a. B2 V1 P7 b( @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5 E, _/ ~; E! S6 q9 G& U$ V$ y" W

5 \1 ~3 b. `3 Q/ G# ?7 vAh, you made a mistake: it should be:) J( ^ V( O E3 b* X
9 v: l b* p' g$ X3 P) |0 J
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:
# \" G4 x2 }3 A9 WYes, there should not be " ' " after "f" and "g", it is a typo. :D" a: [* J" u2 M* [8 c, J1 r: q5 A
) R9 @3 N+ G6 o& y% p
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:' f6 _, v$ P$ E0 s! U) ^- {1 N
! a. X: Y) H6 t' A, P+ N$ q9 ~3 X' M4 p
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。: R6 A. m- F" s1 b* h
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 l. k* D$ c3 T
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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