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 20:41:4 H6 }3 @( d. k1 \9 `, T
请教大家一道微分题，多谢！
; y  I% |6 O' P7 V$ n5 b- i; B! N5 F' b
d [(a+bx) ] /dx = -kc +s
& `8 d, x( d& R- n' i8 \( d  Uwhere: only x and c are unknown, others are all known,  requiire c = ?" x# i/ O9 M# D9 q( q' I) T
2 n& q3 H, _  Y5 S9 L0 r. d
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:
  Z" [! j8 C7 v5 Tit is too easy:% s+ D) [+ B% |- Y$ i

! W: V9 `9 f  L- Q# M( a  N: _d(a+bx)/dx = b6 B* r7 Y& \* \" D
' A  p/ o% a+ K# v( L- q; T
so
! Y) u, o9 V' O6 P1 s; \2 r2 j9 b; s. g
b=-kc+s7 |6 M. Q7 _; `% ~! I& ]
" u, u+ D  q* O; A5 X
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:
2 U* y- Q' V6 M; G( u9 L请教大家一道微分题，多谢！
1 c$ K1 U+ M, U. r/ r
+ F* i; m1 A  C1 nd [(a+bx) ] /dx = -kc +s
0 m2 x& V' T* rwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
9 _" Y1 Q. X* g( X3 s6 Q/ A3 T- E% V1 f9 y1 b! t
多谢了！+ `% }; q$ N( a( k% g, ^, p! w
3 `  l/ M+ m& y' O; f/ m! P0 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:
6 C% W) ~/ U* esorry, did you post another question? your statement is still not clear.
1 l3 ?, U$ N" Y& Y! n! ~- m$ v" [
) i8 `2 C; m7 R A' D: f$ v2 @) K9 Ra, b, s, k are constants? c means c(x) ?. h' ~& V8 @$ v

4 T v0 K e& f8 u; M2 @8 b7 qplease 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:& C- o( H3 y8 w% ]
对不起，写错了，忘了变量c,应该是
1 W0 h. a& `- w* V- J1 bd [(a+bx) c] /dx = -kc +s
3 x, B6 P$ C$ y" E4 N+ [/ u, U* w
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:
. w$ K/ j; O; ]/ l5 |do you mean it is:
& c. r+ H; s" ]/ b
6 K6 }, f, j4 ?3 {d { (a+bx) C(x)}/dx = -k C(x) + s
. f- [) B$ J( k& H5 R- J' y: c3 q1 ~: s5 F0 W d; b/ l- b4 Z, A
where a, b, s are constants, you want to know what is C(x) ?
0 w* ~3 s( T/ n$ c7 S t
& P+ \( k; b6 X8 q' _9 ]& Zif 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:
1 K) e! p3 g6 i6 w1 f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:( b# g6 G' V% w, K% }8 a% D" L
9楼的答案有点问题吧。, X$ t" g* Z4 ^1 U) m. [- @
您说：1 X3 b1 @' h' L- j
: t8 Q$ s. a: \3 ~8 Z: l
d{(a+bx)*C(x)}/dx =-k C(x) + s5 s) T- n) _' o$ g8 K  R" u+ }
so:
, F) n- r, u! {- s# h' t  jbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
6 q2 L0 \& }' q9 {& M
+ j$ x" n, w1 Z+ n0 h0 p6 ^也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx  |) _4 s( c; d; C* ]
但这是不正确的5 S5 M9 u& c$ z- Z& S' q* p7 O
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
9 R* M8 S2 c0 I6 l....
" `5 |( w& }% i$ Ld{(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:
/ N H$ A3 e8 ?7 A4 w这样算混淆了微分和导数的概念。9 x* @. A& K8 C9 }
- U+ N5 j9 _. O$ |# {; Z
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
1 P: N6 E' b- U5 O q* P. Y& Y/ X; f! I4 v# w- z3 C! b, f! 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:
: F4 T1 J0 w& [) y- W$ w- S s这样算混淆了微分和导数的概念。

0 _& h" Y5 H/ h' S0 Zd(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 c. o* h( G- W9 z$ W: B8 e/ }" R' E) z# B& z6 P
Ah, you made a mistake: it should be:
" b- }1 A6 B' u9 r! ]
% s1 ~8 w! s" X: s1 v2 K) }" E( Pd(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:
/ ?9 q) z3 X1 F5 K( e6 X7 hYes, there should not be " ' " after "f" and "g", it is a typo. :D
) o {2 V2 c2 \4 t8 w
% n) v) t$ f! a p6 fBut 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:
- v$ c+ j3 x. e& V1 G! e8 e2 G/ R* B, q: k3 R& W5 Z
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。5 t5 y* F9 l6 U$ w
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:
- c* u( d: R6 Z6 d$ A D1 O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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