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 20:41:3 G" N0 l/ K: H: Y
请教大家一道微分题，多谢！
: T2 l+ A+ w. n8 b9 f- R! \! s! O% t  u- ?! z4 }
d [(a+bx) ] /dx = -kc +s ; l8 c9 P2 r8 _  ]
where: only x and c are unknown, others are all known,  requiire c = ?
. }1 F( C) q9 M% o1 T  A: P! Y7 a4 e7 B; W7 j; x/ [
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:
- |: d2 Y3 n3 d# q2 _; uit is too easy:
: u: \: J5 F7 I" [
# W2 G4 e' T0 }' g6 C* r  r6 r, ?d(a+bx)/dx = b! ~, d! s. f) X% i$ A* ~" d
* W9 N! K/ z3 L/ s
so
+ |" \$ W5 r3 r" k
" L$ `$ `% _7 C. xb=-kc+s- @; \( \/ d$ }  {6 n3 G! O
6 V6 e' J3 Q8 a9 v# @2 m4 ^/ s
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:
0 r2 y3 A, R' X9 S9 ]0 n# Y. `! f请教大家一道微分题，多谢！
( F% }. h% s/ V; W) R6 [
) _- q) p' o5 Z' Nd [(a+bx) ] /dx = -kc +s
# z4 S" |8 T: a) n+ Wwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
7 S. M1 N# F7 k1 a" d
0 E, J; X3 U. B; f/ ?+ |& O多谢了！
9 d7 e6 K4 \% h( W4 n% c
$ q2 P/ U) N5 E+ R) h3 i+ C[ 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:
7 @6 N4 `& `5 `6 u2 N2 vsorry, did you post another question? your statement is still not clear.  E: j0 J; [: x/ ?7 b4 h) F

; J% b3 Q% w3 t+ ra, b, s, k are  constants? c means c(x) ?
5 u( L) ~7 c# _' [2 e+ I* x* f  _. V& U
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:
$ ]2 b. j/ I* e对不起，写错了，忘了变量c,应该是% y1 I/ G T) y8 s) i/ K% m/ d
d [(a+bx) c] /dx = -kc +s
. X5 j8 Q4 U5 R) k* ^' J3 E0 b: Y, Y) e
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:
- X. W% |% v9 }- H5 sdo you mean it is:* S* C" r- x% z
5 A0 a# y7 B$ E0 L+ [
d { (a+bx) C(x)}/dx = -k C(x) + s
; z" ~; l: h6 ?9 t0 R# X0 _- q8 Q! r+ b7 L( I T. Q5 q
where a, b, s are constants, you want to know what is C(x) ?
4 T& c4 e- n: X4 I+ H
# f& v q* d! _/ ]# g( Bif 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 |( I( @4 G$ D5 M
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:
  t2 u/ ?6 @! p9楼的答案有点问题吧。( h. {% j2 E( z# |
您说：
$ k1 K% ]. ?; ?! C! k8 `# G8 [1 |3 g; a! {
d{(a+bx)*C(x)}/dx =-k C(x) + s
5 d! W1 m/ U) \7 Sso:  _# s" }  I9 Z, L" B
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
$ W- x, ]& o) W8 V
5 a" ]( R$ V5 @; \也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
1 L7 T1 O: d: E( {但这是不正确的
$ e7 m3 x+ w& v. Ld{(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5 L) l) C, w! H# }& m5 c' D, H
....
7 [# J5 V$ C' n6 o- R( v6 A5 {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:% t0 A- \ u; V, r# O- F; y7 |6 s: T
这样算混淆了微分和导数的概念。
% H6 ]3 F4 J0 S7 ]* }9 q: y" O# j+ v" Z% d
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
# H; P) |$ H: Z9 \
% ?1 `# e6 L. x+ h g% M) qIf 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:
$ Z4 _, {9 y( C这样算混淆了微分和导数的概念。
% n- N2 Q" Y2 S: x9 z. h; P$ Z0 _; O r: u5 n
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 \* d0 l9 e; v5 q+ ?
# v/ d0 V+ S$ {; ?" t9 I( p# J8 l8 q% TAh, you made a mistake: it should be:; s& N* K4 w. {0 O3 W$ [1 u
& U0 i: j! y& a' e
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 20:51:3 D& q6 H9 E8 P) R1 _1 ?+ V4 `
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
# R0 @; q. W- h% Q; a7 d& {( @5 @6 ~; w. t/ b) b* 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:
: U" H1 k1 A" R- c
" `% S3 x8 Y* f# S! z( i) V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。1 G- k9 M) k' S' a+ _ m; N
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:
2 B, |) s8 |' e! R/ M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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