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:
6 w$ e' x$ n* F2 T' }请教大家一道微分题，多谢！
2 F5 G3 D" L2 U  V& i9 Y" B' U2 g! C7 ^% J( N' U$ O6 f
d [(a+bx) ] /dx = -kc +s
9 M$ n, q+ d: Q- M& w+ z- ?where: only x and c are unknown, others are all known,  requiire c = ?4 @: u1 W8 v4 b) u9 [
( m' ?  g+ P4 g8 k
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:" F) J6 i6 t( M; S
it is too easy:
6 F+ ~5 D# ?' B! {1 c( N
* ^1 d. W6 B6 q( @" H1 g) Yd(a+bx)/dx = b2 E4 _1 r8 u" e4 D
0 e9 b/ N; q8 F
so1 U; I2 V5 a6 E0 j$ g

" |, V2 a, [9 b5 A* V4 ]2 cb=-kc+s) j* o" ?( a% L; x" O" \( r) Y

! R; z L- K4 |% D6 Gfinally: 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:. x, I/ [, v4 U. F" Q8 o! a9 |
请教大家一道微分题，多谢！  G: f( Y' g5 n
( O. o3 R" s) e
d [(a+bx) ] /dx = -kc +s 4 x0 G: \! |3 @% t
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
  }# i' \' d! I3 ~6 K$ R
, N! N/ u3 A+ a多谢了！
( j: s* u% O8 s5 J" o; O. c# h2 ?
[ 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:. y3 S/ a( }3 j( J$ X9 Z0 a' N! L
sorry, did you post another question? your statement is still not clear.
* v  N' a$ c7 W3 g1 m3 c
. A; P% `( k+ ^- s0 za, b, s, k are  constants? c means c(x) ?  U9 _! V/ N" ^; c- G) z
2 R. e9 |3 }: K% `: v# D  C
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:
n. {! \4 f ?! V% {对不起，写错了，忘了变量c,应该是5 W' p/ N& f" B! R8 E4 w) j
d [(a+bx) c] /dx = -kc +s 5 f1 y' Z- g( N& r G: T; l! ~

# Q6 a1 l* x0 `1 k2 h* V0 s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:" [4 W  z, H; ~' B! v  x
do you mean it is:1 {* q$ e% u/ d# k& t' C
3 x7 W( d* Z5 z8 {
d { (a+bx) C(x)}/dx = -k C(x) + s' {2 C9 ~9 O* S  I5 K

  I0 L3 I* v0 [7 l5 W0 ]where a, b, s are constants, you want to know what is C(x) ?/ C  X6 Y7 i3 I( r5 [+ ?
5 t" S9 v: h5 V8 {% q0 T0 e- d. e
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 14:33:7 _# |$ b$ @4 r7 ?- T" 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:, M( {. o. L# `1 c9 _8 H
9楼的答案有点问题吧。
0 \" B! p' t6 C, W5 J; t3 D1 m% j# R您说：, t7 M" j5 F" q8 P
7 r  D1 \4 w+ f5 S! w) e
d{(a+bx)*C(x)}/dx =-k C(x) + s
# W: P% Z$ r5 Rso:
8 K% u2 d9 H2 b- t" A) ybC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
4 M( b5 B) o* ]: U) L8 b5 C9 b5 q+ N. v& Y' D, u# V" p5 W; k3 O
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx  o/ M9 g. v8 y9 x- [8 {
但这是不正确的, d$ G; i% W% H- ?$ `9 R9 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
4 l8 ?( E5 Q1 H; l9 L- q; m....8 J4 \$ N& h% f! U* G% V
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:
6 {- ?. M$ o& l! A这样算混淆了微分和导数的概念。 a/ H4 g8 j/ Q5 p3 \9 |' R

, n" x# z( E% td(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 o/ l5 r! ?6 ?: h' _; s* }5 p6 P- i0 ]# W9 j" D/ C
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:& i4 Q: ?; Z# u
这样算混淆了微分和导数的概念。0 y7 O4 ~% ]% ^# e

- K* d4 C) L2 F% ]$ J' H" B/ Od(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
. r7 n( G1 ]2 @# U% l% L: s) B$ s0 s8 Z6 j1 f* b0 S
Ah, you made a mistake: it should be:
, J% x9 i( K$ D; o; {! L* y# Z# F& y* c
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:
- g, I! S3 ^' X$ Q' p! }Yes, there should not be " ' " after "f" and "g", it is a typo. :D
3 j6 H4 T: R' Y0 @% A$ m; I0 T9 _1 [/ b
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:5 E; g4 t0 K: ]1 X1 a7 ?

k% J; y4 u9 e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 P2 I; h F6 D, i8 W& c6 C' 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:
9 ~9 y7 ]& o; J2 Z) t- j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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