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:
5 t5 @6 h4 j- a2 `2 Q( N请教大家一道微分题，多谢！
9 `; r( W0 u+ H4 @9 \0 b3 ^8 A3 w: n: ~0 t B" o, V
d [(a+bx) ] /dx = -kc +s
# ?8 a+ ?3 |& j6 T1 ?where: only x and c are unknown, others are all known, requiire c = ?
; Z5 |; {, ~& @+ T. w5 D7 }
7 t& `% y8 k4 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:. T1 ~+ g0 q' ?! l7 `- b/ l8 d
it is too easy:$ B- O- Y$ n; z% V' S$ ~
7 l/ h7 n: j7 J3 S. D
d(a+bx)/dx = b2 _1 [; }$ G1 Q3 n- k, @8 M
7 K3 M6 g' e9 D/ S2 j" y
so
8 B' W  s! }3 s8 N- ^& T$ n3 l
/ {  Y  t0 h6 H8 ^b=-kc+s
  }  |5 \' t/ P6 L7 P0 E8 N2 e6 S9 Y/ L$ ?3 }! H' S5 S7 G
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:
, Z. r  k6 q8 M5 l5 S* g2 y" l请教大家一道微分题，多谢！
9 W2 ?: o) G3 |+ i8 Q% E9 P: j0 C1 P/ Q  C- @
d [(a+bx) ] /dx = -kc +s
% r4 f% U5 T4 c; O4 c3 ~! m2 Awhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?( s8 X' g& f6 a3 a% Y

8 Q4 t% g0 C5 n( j+ L多谢了！
- R! v' B4 `* D: Y/ H1 n* \$ N
! ^% s' i8 \% J2 @[ 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:
sorry, did you post another question? your statement is still not clear.0 k! d' ~- i/ O% |6 J& {

: O- H7 Q% R2 @5 P$ Y% v/ ha, b, s, k are constants? c means c(x) ?
( V# m+ Z7 @% }1 @, K2 y4 t: u2 ]- E, G( O! a% P
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 C, z0 }; S' d. b9 {1 l
对不起，写错了，忘了变量c,应该是
( j( `" x% }- A" {+ H, wd [(a+bx) c] /dx = -kc +s 5 Y* x7 e4 s' Q

5 ~+ P/ y1 D Y9 { o+ i0 V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:
$ I+ K/ v' K0 \) {4 _2 I9 s* [2 _6 ldo you mean it is:
; ~( a; t/ y1 \; B8 B! F4 I4 [$ i$ B Y& C% l2 n( i
d { (a+bx) C(x)}/dx = -k C(x) + s& i. p9 e2 B) L6 o ?

- G9 I. o# n$ V0 `where a, b, s are constants, you want to know what is C(x) ?
. A" l* s3 {& S0 X* U0 ?9 w" b% t, {$ i4 K3 ^# S, B
if 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:
; \7 o/ r7 U5 w/ H* |1 }& h( v' c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:
$ [) Q( @) ]$ @$ x& c9楼的答案有点问题吧。" `- m8 `' T/ a* Q7 {
您说：, ?; H9 z* V8 X7 `
* t: h! n/ g3 P2 v# a/ R4 q) h
d{(a+bx)*C(x)}/dx =-k C(x) + s" n1 T$ Z: F/ X
so:
0 }$ A; ?$ g5 u) ?( G3 R, ]bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
- s! w3 S4 U' h
* {. i, q l( s/ @; n A7 E9 J也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx! E f k! U+ e* z
但这是不正确的3 m, g: R' T! G7 v7 e8 u) C
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
+ L. ~- E. l: o$ F) M9 S: @....
0 [) ]* G0 {. ^& `/ Q' |6 q7 Yd{(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 20:30:1 G' Z1 N6 j5 V) X8 y9 T( D' s9 C' E
这样算混淆了微分和导数的概念。 ]2 t6 A+ d$ }" ~4 M
# X* l7 e: ~5 Y E) B. r; {" X
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, }3 ]* c7 a0 P4 ]+ y6 O

$ |; S5 O6 b: m' K% U( e5 eIf 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 |' W+ ^, Y2 ^' }6 ^4 o8 G
这样算混淆了微分和导数的概念。- S; R' u5 E) N4 I
, J+ B/ ^0 T4 e+ H5 J% @
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9 l0 K! Q L* l; ^
& n L' J: v k$ P8 A
Ah, you made a mistake: it should be:3 }5 N* d7 r+ a# @

/ ^# s5 [' g* g2 j2 C( 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:
0 v8 F }" U8 T; @Yes, there should not be " ' " after "f" and "g", it is a typo. :D1 J" N" Y/ G. s; e! y" j

! S1 ~- p' b$ sBut 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:8 \2 Z/ v0 F! G2 }
% \0 N5 a9 U$ _ M3 M) M! ~! ?$ P. t
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。( C' o/ P$ a3 P9 B8 J3 |) C" y! {7 f
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:
! M% c2 n, J4 t( [9 w6 n: R i& d$ 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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