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:
0 }- M+ S% M3 A* Y P p5 A* f请教大家一道微分题，多谢！7 x) W5 B8 [; t

9 W6 V' R0 O- C3 M* vd [(a+bx) ] /dx = -kc +s 6 t/ I! p6 _, q' k+ F6 F
where: only x and c are unknown, others are all known, requiire c = ?! }9 {5 X# t% T/ v6 K
# i& s& k( F6 Y1 }7 [/ L5 g$ _
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:
8 [# L4 e1 J/ z8 Z% c* Vit is too easy:
8 ^5 r  g0 ^" K
9 z* d1 c) v, a; `: o7 H' ad(a+bx)/dx = b! L; M+ w+ I1 j% m4 K7 y
0 }4 z7 e2 |' E& a4 F
so
6 z- m' e1 Y3 W5 W: q, y: x, i& {+ `# u  W8 D( R; O
b=-kc+s! v8 o6 r( P/ i8 @" [
( u; k( S( j! Y: H0 I6 N
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 20:41:7 B/ Q5 j4 c# e! j
请教大家一道微分题，多谢！
/ A4 o e, V7 k# P" e1 b6 y2 \( I4 q& F" n" b
d [(a+bx) ] /dx = -kc +s
+ a& Z& W- m7 p9 Y' `where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?: x: ` H- d \( ]1 ~8 N
2 S% I6 l2 f! u1 e9 O7 e) V
多谢了！
% Z) o0 ^0 M" Y. `4 @/ \! @8 c/ W; t8 V: p) v, J6 v
[ 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:
; I: ^4 H! g$ p2 ]; Z2 [1 Gsorry, did you post another question? your statement is still not clear.
, w* N7 r' }( ?3 R g" K
( O0 ^' B- ~" B7 T4 _2 _a, b, s, k are constants? c means c(x) ?
/ g1 ]4 n( w; |% \2 c# f" T6 w. [/ |) Y& f) F5 G7 F
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:
! E# k! \. P) k" ?$ C' y6 u对不起，写错了，忘了变量c,应该是$ v c) u1 {) W8 Y: o
d [(a+bx) c] /dx = -kc +s 7 Z% s* `( o$ E$ v7 j& ?; R

+ ~4 T1 m) R; X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:
+ T# t/ w+ s. T! r+ H1 i" k" y2 ?! fdo you mean it is:$ m3 n7 x- @: f/ I! [9 o) K
4 R6 L2 P! C, P; |/ A0 E
d { (a+bx) C(x)}/dx = -k C(x) + s5 a# L5 u7 @9 k- C) Y5 g! A0 u
* A! l+ q; i- H/ u% |1 N# ~
where a, b, s are constants, you want to know what is C(x) ?
" \- a3 z" j8 T* z' I {
0 O. \( h% X5 {% U- R& M! d0 E( W. S: 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 14:33:8 I) u* }2 K* V( [; ]
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 ]8 S7 J3 i, d. D7 M) C9楼的答案有点问题吧。
; k$ G1 n/ a+ p" @您说：5 P4 k; ^4 h+ J. j. Y7 t
3 ^, |! N& k; V' {% _* _
d{(a+bx)*C(x)}/dx =-k C(x) + s7 R% U% p" I2 a" }: J$ ~+ i9 X" l
so:
5 @9 M- i% @7 V/ QbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
9 q2 d) q1 `! F8 C, q) U. T' F( j% z5 k
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
9 W9 t% J2 ]& ?( o0 n7 I! }* m7 C但这是不正确的
, q1 ^5 d/ W" T6 }6 [7 Kd{(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
( ~2 R2 q4 {. J N; R( [* O1 S- ]....
: e3 t" i; D$ B! q+ ^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:
" c* P8 d7 `9 g& x9 J) A2 _5 B这样算混淆了微分和导数的概念。
5 B) x1 }/ f. R, h
3 S# C* D; G' }* Y' T: pd(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
0 X6 V- N3 a% d: x j2 x" }% x% o
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:
0 i, K( k7 j3 k0 T6 x5 |这样算混淆了微分和导数的概念。  v' Q6 Y$ v, ^& w: D

; u( O0 p, u' u1 P$ b2 t# p. ^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 g7 f( ~& c, {( O$ \) f2 _7 w  t  p/ y, T; V$ f5 O7 E& [
Ah, you made a mistake: it should be:. U0 E/ J  u3 M7 Z8 z9 ]8 B# u4 s4 ^
& u" [. z0 L" t
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:; K5 ~+ ~' G) t4 [ E
Yes, there should not be " ' " after "f" and "g", it is a typo. :D2 x# H4 g5 N: R8 v2 H

2 k7 U0 ]' X5 G: ? ?* l9 P$ aBut 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:7 u, b. S6 ~& v' d2 n! i; M6 b) r6 P. B

( n1 a8 X9 e7 H* l1 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。3 ]( E0 A$ P4 F5 T$ ?8 }
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 23:24:7 s9 o! Z: C8 L- [6 Y* i
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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