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 v9 w8 }' D' W0 p. o! j; t( I ~+ A
请教大家一道微分题，多谢！
) B6 A/ a: I1 y* W( h) i) Z+ X! t; _% U: k& w( j4 _
d [(a+bx) ] /dx = -kc +s
$ R4 ~% l2 @/ L8 ^- Vwhere: only x and c are unknown, others are all known, requiire c = ?
0 X7 A. \- `; `# h5 E( p$ q
0 a8 |( Z. ?! c! Z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:' O$ h" C" Y0 T
it is too easy:
3 ]/ P9 O- \3 I0 ?$ J1 _) w2 \+ H% e1 F9 A7 N
d(a+bx)/dx = b/ `3 G0 W1 r1 l3 B& e: N
5 \! E+ A+ Z% X
so
6 k5 j, L2 e' ]7 {4 E- h! G/ A: j6 b9 t, O# R) d: k
b=-kc+s
2 N) |6 X, H# W* |( ^
& t% q8 p- A+ A& M$ G+ D: m( @# ofinally: 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:
- s# i, O+ z: T! I9 S" c请教大家一道微分题，多谢！
( X) m* `8 y% ?/ \
4 w3 C4 y7 |+ X" x9 y1 Gd [(a+bx) ] /dx = -kc +s * @* L/ k% k- U, [: S% R3 q
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
- f) s# }& X" |7 I' i- y( |( J, B& z0 u2 v2 j' U( p
多谢了！
' X% B; h& u, ~& E( p/ c# I1 a, `9 g% J! ]3 [7 d: u' A
[ 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:' C2 r$ ^. B5 j" A: U/ k7 }% o
sorry, did you post another question? your statement is still not clear.
* Y6 `  o% L' L% @  m- g+ p& U
, X8 b/ d# F# P$ i7 ha, b, s, k are  constants? c means c(x) ?
+ j) c1 W3 j- |; R( s  ]. E% A4 I" d. m/ M/ I
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 12:30:3 o8 d5 Q4 e2 m5 `. }7 V0 a$ L5 x
对不起，写错了，忘了变量c,应该是
. p- T- }9 o$ e; ^- B! g5 b6 ud [(a+bx) c] /dx = -kc +s
( E! {: [, l7 Z% H3 t- u! n& i9 M+ _; G1 g; K
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:
- `5 w- S" Z t3 C$ C* e& pdo you mean it is:7 c; H! K$ d5 ~ {. F$ b

0 }( M5 _4 l" F6 s9 qd { (a+bx) C(x)}/dx = -k C(x) + s
" D0 J# O7 Z' x
! }2 \ U. y0 a, Nwhere a, b, s are constants, you want to know what is C(x) ?4 f* b9 a V9 n4 E$ l

/ D, g9 v7 w! sif 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:# v0 v' r0 k( W* J% V5 H# T
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:
0 a+ b3 @8 Z8 B% z- u2 `- n5 D9楼的答案有点问题吧。
您说：% X$ W r2 h# l! N
& J! E, Q' K7 M8 f
d{(a+bx)*C(x)}/dx =-k C(x) + s
( a( a! Z' d2 {- T8 Q# Eso:
" ]) [+ }4 W. \% p& WbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
: `, o) U8 ?5 N) M2 h$ f
! {' I' p4 g& H# q# j5 R" H也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx+ I+ r6 w# g9 V6 z
但这是不正确的% L5 _6 R2 U, X1 `! m% w# n+ h
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+ m1 ?: e6 k: t
....: P) A& I5 | _2 Y
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:+ ]8 d2 A O+ F/ p1 J: b: j# J! R
这样算混淆了微分和导数的概念。
0 Z( T! m& U w" j. ^) R0 _* c# P- m& e" H, q2 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
5 G3 t; }3 ?# g2 r) X$ V
4 f& a# j" l8 O4 a0 gIf 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 _- C  t; w( j
这样算混淆了微分和导数的概念。
9 Y/ D+ ^7 ?6 W: m
' y, M) E2 Y& ~( M2 m% T5 F% Ed(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
' D1 `: c! F* C# r& s: e+ s4 K  l: ]4 X8 I
Ah, you made a mistake: it should be:. a# c0 a1 o( V% J0 s! W) A# b

; v/ a! Y  u0 `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:, n9 `" K4 N$ U
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
% l: b2 D5 r5 { ^+ @
( d+ h! d5 C0 ^- L& c9 n! 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:
9 t0 Q1 p1 W, V# ~# X$ E/ K2 P
: x7 y! t [+ N6 y% y6 }* A* u$ A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。+ k) X' t2 ^% V/ f7 d1 l
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 23:24:3 T5 K! x( I# E' B
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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