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:- m6 u2 f# \; \* x
请教大家一道微分题，多谢！% q) `/ @8 b! T0 ^7 P

* G# x+ O$ j, u* b( _, Q% Z' Ld [(a+bx) ] /dx = -kc +s
5 J7 J! J) s* X- {) C# gwhere: only x and c are unknown, others are all known, requiire c = ?* t+ w, X4 m2 M3 y2 e3 w
' X. w* E! ~! r* Q: ^2 s0 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:; ?; J; O' o6 D; X! w0 Y' i6 D
it is too easy:
4 P6 y" O6 n% R9 O% h
( w& U; y* G) q8 n: L6 g! N5 q# Cd(a+bx)/dx = b) [1 G$ N+ E4 I# u; r0 Q
5 w) H5 [- m: U: X& l& H
so% W9 Z5 w# {. t/ V/ U. o3 C

# ]% q# l8 @. Q5 nb=-kc+s3 i) A* N1 N! c
4 q( [! ~2 U2 ~5 E
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:% r3 e8 ^6 n, }# w* e! m3 K
请教大家一道微分题，多谢！/ A9 \% [ d( E& L3 o, r
, t2 l, j# G) |4 M- H& [/ V/ f
d [(a+bx) ] /dx = -kc +s
. f! f9 J. s- a9 ^where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?0 Q# {$ G3 x k" q8 y& w

1 a0 K5 k' }/ c; \: f1 o$ W多谢了！0 z. q# Z3 P( E

4 |; u( C2 i0 G( [9 q8 L& X[ 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 f0 k. j; O5 Usorry, did you post another question? your statement is still not clear.
9 n3 b4 j' {" G, e! N. a' L/ v7 e; M( Y# [
a, b, s, k are constants? c means c(x) ?
; U6 s- ]! F5 A. t. A
; B7 N* @" g+ }/ _1 Qplease 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:
- S$ t% D& r8 q! [/ D5 J8 ?1 _0 h对不起，写错了，忘了变量c,应该是8 M- B/ x4 L4 I, s0 t' k
d [(a+bx) c] /dx = -kc +s
; }% D- L3 `, O# u$ W$ N/ y7 D6 T1 ^- w$ K# @) n& V
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:
, h2 X; X5 G( o3 ?3 [do you mean it is:
: K- y" I: S! x' b
5 o: _0 P8 B% R k# J. ed { (a+bx) C(x)}/dx = -k C(x) + s% k" p# y4 B+ x# B) Q9 N9 e: @
7 |' }8 @/ W7 n+ O, X" |
where a, b, s are constants, you want to know what is C(x) ?
/ \' A8 _( I& G" ?! k6 e/ A+ L3 d1 h6 a3 ?, \$ x( a, j0 r
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 14:33:5 ^" q! o1 f1 n$ \2 H0 e/ r/ w
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 19:34:9 l+ k. j, G) \( p# y* [
9楼的答案有点问题吧。
' \, u. X; Z. T您说：7 I" ?' B) m) o3 j% G

1 m4 N  T/ A9 {/ A  a2 P4 X+ vd{(a+bx)*C(x)}/dx =-k C(x) + s
. k& K( T  y3 @: Qso:7 q6 T5 k) g3 X0 p/ W
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s( h7 t+ U+ p: h& b9 s5 l9 r7 _
7 k2 f+ L5 B* N8 u  l
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx' t1 X4 ^: @# Z4 ^+ s- X
但这是不正确的
% P( E, M: ]& S& W8 ad{(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; N1 Q `$ j. {" ^& {$ R% I0 Z
....
% y3 Y9 K8 \2 {. }/ cd{(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:& H6 ~. C8 Z7 T, \( u
这样算混淆了微分和导数的概念。
: d% X6 Q# O6 l
3 k& X j, g6 i# H% N# sd(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 {4 `! P; ~6 ?
2 h) G! u2 E1 w& o/ _" YIf 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:4 X! R1 N: Z, f9 n9 L8 j" l( c- C) O
这样算混淆了微分和导数的概念。( m4 }0 y/ s& i

8 ~4 T% v& Q8 a+ P0 Pd(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/ g5 j$ e) K9 d9 |1 z- I
2 M1 E: ~) L' t3 t7 }" q0 G
Ah, you made a mistake: it should be:
1 {+ v3 J8 w5 B
+ V( ]: ~ r; ^" ?. G8 { Pd(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:" V2 R! A9 o1 g4 G/ n
Yes, there should not be " ' " after "f" and "g", it is a typo. :D) a2 C6 f9 P- H% H5 x8 b( F, ^

6 T3 B) ?7 R1 k/ }/ Y5 J- c0 r) EBut 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:
; n9 Z6 o8 Y! C2 Y' R& w8 p0 @3 d0 i Q+ ^- @- A* G
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。, |2 [8 E5 q" T
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 w% A. y) R% x4 V7 z: S! h) 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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