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:
2 w* F, j0 S# A: |) q, r+ P请教大家一道微分题，多谢！% b1 l7 d6 Z5 Y, K$ g7 Z
0 K% U7 d3 Z: x: L, b
d [(a+bx) ] /dx = -kc +s ; }3 S: f- K. T" u! l/ c4 k
where: only x and c are unknown, others are all known, requiire c = ?
0 n9 D! Y4 Q7 \7 V& n) k3 U/ d2 w% W7 Y
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:( p# \$ r  C/ n) r
it is too easy:
! ^3 l4 I5 n! s
$ Z' w% j, ?- x+ }# Kd(a+bx)/dx = b1 U0 P4 X* f# g* T. b% U  E
) {' R8 ^$ M4 O/ c
so
/ A/ ^5 E* n$ Z7 v7 X" l( O# I- \% ~; |, s% [
b=-kc+s  Z# ?$ m1 J9 W# W* ?

, I2 I: ?# _2 h+ R( j: w; \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:: Z# g/ p, m" D4 w
请教大家一道微分题，多谢！
$ z/ X; ^  v( [& }( [5 u5 c- _$ `7 D/ L$ D- ?, N% f5 u
d [(a+bx) ] /dx = -kc +s 7 A  Y8 r, R- T- q
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
" h! W6 V5 B8 c; U
) z* T' L* }" v! |# A/ I. A! `多谢了！
# c4 @6 l4 x$ c5 X8 W  [2 Z; m
3 v. V/ G& Y' e* {/ E[ 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:! X7 Y, L, e* F* H$ {6 Z8 S# R
sorry, did you post another question? your statement is still not clear.
; q5 t4 E  j" V) ?% d3 _+ y3 K7 q) d0 W8 b3 K  u0 s
a, b, s, k are  constants? c means c(x) ?
( x4 x7 H) n  z
4 S2 |8 B8 D1 G5 o: k8 ^6 [* Cplease 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:
2 b4 [$ B! w- ?6 G5 C, [8 @/ E! A对不起，写错了，忘了变量c,应该是4 c# }3 i& @ o$ p2 m
d [(a+bx) c] /dx = -kc +s 1 I4 y7 k! i; B2 T0 t
, z; D* v9 u3 \& M# B3 n& B8 v# 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:! _' S6 n. ~! n: W* P b h8 l: R" ^
do you mean it is:3 f6 @( k# n: d: v- X

7 y; _5 Z# l" P$ C0 S, Q& qd { (a+bx) C(x)}/dx = -k C(x) + s
, C! d: e9 O4 w' W' n: a) v* A
4 w6 y. }' ?! b n" Uwhere a, b, s are constants, you want to know what is C(x) ?$ B* M4 K' _/ s8 Z5 \4 d* w- \& e" x, g
2 l( V/ G0 K4 n* K4 r" Q
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:& V* X" G: |, C. r4 d
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:
" L- x1 E+ ~) }( g: h( g9楼的答案有点问题吧。$ Q! v8 O. B3 }7 k
您说：4 u% f. I+ k; D
0 |: m  s" ~( K7 }
d{(a+bx)*C(x)}/dx =-k C(x) + s6 a" y4 t1 j) N) A. ]7 p% G: _/ }
so:
# |7 M* ^5 X3 i7 hbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
3 P  P4 ?: U1 ~1 f  J5 L3 W+ \7 L
/ K: L6 R; ~9 B; L" r4 ^/ U也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
8 ^! J* x' `& s但这是不正确的" k% m( R( J+ H4 p
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0 e5 a6 U! l) u2 \! e& H p
....3 R! v: x$ [- W& ~/ o" T
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:
1 H' n1 E4 \7 e# C这样算混淆了微分和导数的概念。
& }& p. G" o$ Z: U) D9 A6 D4 T5 M% C$ N5 Z
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- S1 _7 D7 S" G( s$ g$ A

2 g: P: l& {$ T9 r! o0 k; XIf 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:! U8 ~" Z- j. T0 C9 A2 q( \9 ?4 v
这样算混淆了微分和导数的概念。
$ u! w2 u- `9 R% {# b) T; s& j( `/ K( R; s, ]* j
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6 y% @, J: E7 V9 H% _
* }7 Y5 n5 W6 c
Ah, you made a mistake: it should be:
5 l, v7 I x ~+ U) X& O3 v+ Y3 @1 w! Q4 R) q% ?! J/ k
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:: e: I" k; L5 ~
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
. C" I# G" i0 C: N. z) y; W7 F+ R, @1 f) l0 \+ 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:
/ \& g8 u3 u( K# Y/ Q" _8 m% E$ s8 Y$ D5 v3 I
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。
- f8 \/ U- |3 H. B- G$ E( i/ d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 g1 x, i! t7 K1 k8 k A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。

账号		自动登录	找回密码
密码			注册