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:% h3 H* d3 }1 N) @9 U* `2 [3 i
请教大家一道微分题，多谢！
- ~2 J- u7 }  E9 f) s7 K. H' H% u$ L; L: F% a
d [(a+bx) ] /dx = -kc +s # Z, H2 z# U$ S/ y* P
where: only x and c are unknown, others are all known,  requiire c = ?% g* Z0 S& i0 Y: c
3 W0 W2 u9 o, Q$ \& v  }
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:
. S( u+ T% i. r# l' z' I  n! Vit is too easy:' l$ h. Z1 t; G% B7 K- m4 f0 ]( W7 l

. z5 ^  C8 I: B/ \d(a+bx)/dx = b
  b6 {! v6 H$ s' o5 z$ e
5 j: d0 @. \4 m& Q4 ]so
* {. v7 Q( Y( Y
: U1 `. C: t& V3 f2 x! K& n+ Mb=-kc+s# S7 L7 `1 w  _1 p
/ X5 x& E! p9 `0 C7 M3 f
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:# Y# e  O9 ]( n
请教大家一道微分题，多谢！* |1 V2 e; H( ~" r% l
; g  h; \' E0 [  @/ o* \- `
d [(a+bx) ] /dx = -kc +s
  B: c" g* P3 [# f. Q1 K% Swhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?- R7 e. t& `! }) J! y4 o% a: H4 U
6 H2 U! n0 C. U8 M  q8 @5 d
多谢了！
% T5 k, H, r! O: u2 s, j
0 t4 U8 D9 ^5 c( k, A7 z( n/ N[ 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:
: e9 V) L) S" f6 Esorry, did you post another question? your statement is still not clear.7 M9 _1 x% s# X& Z0 ?

* _! v% J4 s. }' u3 l Ma, b, s, k are constants? c means c(x) ?9 s) y# r, o' d3 I* O

7 ?" |9 P3 f b- lplease 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 \3 D/ `, g- L$ k5 C# |# n- L对不起，写错了，忘了变量c,应该是
+ e( ], L, k& U# o Y0 od [(a+bx) c] /dx = -kc +s
' D, C/ ^1 U* B+ e1 X# C
4 r* W) }, L" K- ^1 Q5 W2 G0 j% t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:
0 n1 E, `- [1 e4 x3 k, u3 w7 ] kdo you mean it is:
1 s2 ^. v0 ?0 S
& z5 C4 h4 J! G# T4 d, xd { (a+bx) C(x)}/dx = -k C(x) + s% p7 u0 _! f2 v' q& n3 t
( c. }% |. o2 g; z6 ^8 X
where a, b, s are constants, you want to know what is C(x) ?5 D% B' F+ `& D( c# E" h

" r4 j$ n0 i' p; ~3 B* C, wif 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:
: q3 M+ j' e' F! k- q0 }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 19:34:3 {, r/ m. z7 \
9楼的答案有点问题吧。
1 E) T" t2 _% h$ e# J# d, W/ ^您说：$ _, F: S& ^/ Z. U
# e9 q/ B0 S0 B5 j* Y4 p
d{(a+bx)*C(x)}/dx =-k C(x) + s" f$ k; j9 ]# v
so:
+ B% Q4 B0 h0 |bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
# S' i& R4 z1 h! ^4 i. E6 j+ I
0 i9 A: G* }) \也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
8 K1 m+ }( A& F& l# _但这是不正确的6 E! b$ \) Y+ v* L0 A. x1 A
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
, T5 o+ K# C: g% N....
% r3 P# N2 _7 N- Z& x: H- S7 h7 ad{(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 20:30:2 u& k6 ]3 r& s' l6 K8 G* j9 ^! Q
这样算混淆了微分和导数的概念。$ _. s) G- U. y/ S2 V
5 \0 k* A5 p& ]: a" }
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
# j3 W& M- z3 [( c, b: T- j
% ?% Q+ w9 M4 b/ FIf 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:
+ p# o" H. }. W# `; ^, T这样算混淆了微分和导数的概念。
. w& E# N# T9 T* {4 t; l8 w( z0 u( |" S5 S& U% G2 O
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
4 {; |% [$ j# ~- s
$ S! e- E2 w: t. P' w+ oAh, you made a mistake: it should be:
) T. x( h2 W Z8 V5 f
9 ?# U7 x0 P6 M y5 |9 }& G( |& h' Fd(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 p( W, }# n: P6 T
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
) r1 G) A7 |: C/ E/ H
0 D3 [4 o) {% Z: YBut 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 U: [6 t* z: Q- O# j! d! B& n- j: f. D; H$ C& v* e+ _1 ?# S
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。! X, b8 F" ^3 ]
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 {7 ^6 F. G! v3 e6 A3 L r% f
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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