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:/ u3 j- A" T( u: x3 e# v
请教大家一道微分题，多谢！/ r, v1 [8 r3 Q9 g5 q- z
# ^% l' K* n9 l' X
d [(a+bx) ] /dx = -kc +s
3 y' }* l9 O; `where: only x and c are unknown, others are all known, requiire c = ?: f: | A0 o- A. ^ C
' _& h* } [- I: p, t6 S
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:; [- n$ U* U: s: w
it is too easy:$ T0 r2 U5 ]) a; _, c4 y! E3 d
; f- W8 V# z0 i6 i1 G3 X' ]( Q
d(a+bx)/dx = b
* c- L# v4 y# Q- h" C
  b7 x/ _  {2 G$ W6 W& fso  A! ]# N2 k0 j  b8 r
( `. R# R6 ^: U0 s
b=-kc+s+ ~2 H  @' U9 L! q  {1 j3 L
0 @" P8 v; f# C2 Q
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 20:41:8 Y% n0 d* M; a  W4 ?
请教大家一道微分题，多谢！; N% R& [/ z. T  |  ]" j7 F

# w4 R: }1 ^8 r; F3 od [(a+bx) ] /dx = -kc +s
/ f% t" \! ^2 k1 N1 h. k7 S3 bwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?" U5 l- D" v" h# x

# p% Y1 T# T/ L( D! s5 O多谢了！
+ V  h$ O. M+ e* u9 U* X9 j
$ J* _6 n7 i+ N9 m6 f/ m[ 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:, o( L; O9 g2 M% e) l  k
sorry, did you post another question? your statement is still not clear.
' E3 u3 s. U( j& |( G) i  y( |; f% u8 J" S* _# s: k2 Q
a, b, s, k are  constants? c means c(x) ?" H& R& Q# w% [, e! ]6 `/ [6 j

& Z4 h/ o# _* J. Jplease 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 12:30:8 }1 a: r8 f/ t n# q# H' S
对不起，写错了，忘了变量c,应该是
: f. }8 r& s' b5 n# K% h4 Zd [(a+bx) c] /dx = -kc +s
4 e6 G, c9 b) J9 c3 I3 {, ~7 `" }( |$ L% K( C
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:' f, ^/ l+ U- O. }$ d% ]; B+ p4 [
do you mean it is:
3 a$ H/ ]; Z% J8 t$ E
; V2 r/ j( Q0 P' O) S. ~+ i" vd { (a+bx) C(x)}/dx = -k C(x) + s
8 o! w$ j5 S; j) ~) I0 Q3 O3 {
where a, b, s are constants, you want to know what is C(x) ?( I/ A# K& W, K

- C( g2 u$ ?# A1 aif 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:
+ i' f; b/ R% [" [# a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:
( J2 B  C: D+ m/ c- r9楼的答案有点问题吧。' X. Z' F; R9 h1 i; h5 M9 Q& S
您说：! `: j# J  z8 K* t5 X: u3 x

6 o' a. \& h) ]' b! r* d+ s: Dd{(a+bx)*C(x)}/dx =-k C(x) + s$ L$ }- o  d) b, S- d/ y6 x
so:  i% F" N+ b9 w& G- \$ _$ G
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s! ~% Z2 x( N0 }2 \' V

4 ^( ?: v7 ?* V! v3 w8 H+ a! e/ V& C也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx! L8 @1 y" x, k9 U9 T0 K
但这是不正确的
1 e7 B1 n5 s" b0 A. Q/ C- q0 bd{(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
/ }" O! {& F8 ]/ e, x- D# Q& }....2 q0 g) U3 a1 ?" r9 A
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:! U' G: O3 z! C
这样算混淆了微分和导数的概念。
" H! y: B7 a8 @3 q* M8 T! i g1 B4 O2 l% 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2 k5 u1 W4 d! \1 S3 a( h
0 R I6 z1 K4 d5 _4 v. G) M' D
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:
) E% Z V* o3 H# i5 V这样算混淆了微分和导数的概念。
* Z* t- e- K5 s6 Q6 l: c; h- x- B+ h9 ~4 w" E) R$ b- v
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2 Q0 T! B2 l6 y, x! c" X0 u8 s

! O. }! N% {8 A/ f. j9 v$ m; Y, YAh, you made a mistake: it should be:
. J1 R& L% B7 V
/ E( M4 b- v k1 N# Dd(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:
" F% |, Q% Q( L; dYes, there should not be " ' " after "f" and "g", it is a typo. :D
; X) F. P. z+ U( q( M' d8 {/ q! h6 Q# R4 e& g+ D6 r! C: L
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:2 |$ {, p& E# B0 g% g
: n' J. q% ~( Z% y# C$ F4 k
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。
9 L& n2 k. u7 r! ?/ W ]/ 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:
; r# b# e; V" h% \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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