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:
, h. f, O- R7 ]7 x! Y9 G请教大家一道微分题，多谢！1 a& e8 p7 R' u
5 p6 i7 T; C$ u* ?: p6 K2 k
d [(a+bx) ] /dx = -kc +s
. Y9 B: ^2 i; @0 f% swhere: only x and c are unknown, others are all known, requiire c = ?
6 l& y9 d4 |$ I: z1 ^
4 ~2 R/ k0 }* o- h. b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:
9 h0 M& v& z; ]' M0 ^4 N% qit is too easy:2 g+ W6 a& y, Z4 _5 M% A
* }) v* f( N6 _0 O7 g
d(a+bx)/dx = b2 O  V# \0 N7 ~* K4 v. X# E

' h2 Y4 m  d1 t1 Y( vso
% G+ Q* i$ |8 H# U* l; ]& k. Y( T3 g% e  O/ v! H
b=-kc+s2 X+ V# m1 H6 D5 `$ U9 P

" K8 E8 y9 T& b4 @. Yfinally:  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:  C- J9 a& K/ ^; R) \
请教大家一道微分题，多谢！; S9 Z& }8 H8 u# h" F
: \+ Z4 K( |* c" R1 `0 m
d [(a+bx) ] /dx = -kc +s . k1 i  d9 z4 A* P7 u, \$ x" E
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?: C1 M- s3 l' P2 M, A3 Q
: V5 E% H6 \; }. ?0 d. }
多谢了！
# c; C) J; b. w, B) k; A& {+ `+ n7 Q2 b, b' |
[ 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:
9 {9 ]8 ?4 {3 Y% w$ A( k" vsorry, did you post another question? your statement is still not clear.: [+ f% ]; G" t0 a# I4 u

) J( H' C6 r& s9 _" }+ g0 Ba, b, s, k are constants? c means c(x) ? q: t1 F* M" [1 R, q; @0 \% q- l* h

- L2 ^7 z, n3 Q- I }+ C2 Hplease 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 s% Y; Q# q ^
对不起，写错了，忘了变量c,应该是, \" z' ~! ~% G8 F* G7 u9 r
d [(a+bx) c] /dx = -kc +s
- R. l) Q1 I K" E( L6 Q( T0 Z5 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:. s- H* e5 W; N2 [( @
do you mean it is:5 T8 E* C; f$ x" H5 c# z0 y% B
4 d' K+ W# z# Z! E
d { (a+bx) C(x)}/dx = -k C(x) + s; r2 Y9 Y$ g( s- `+ _+ D

$ k. o/ v; g( ^" Swhere a, b, s are constants, you want to know what is C(x) ?0 s) l' o8 ]8 `( h3 |0 |8 ^
; r; \+ i: v: Z0 p$ s) h
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:
. n9 v( t5 |! u+ p, 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:. Y3 |5 M( Y3 j" D
9楼的答案有点问题吧。
! x" `  m2 S3 @6 H8 I/ Z您说：$ C& w# w; ~9 E1 O* Z

; }) e" o* ?) f# ]2 c, X0 V7 |d{(a+bx)*C(x)}/dx =-k C(x) + s9 @- ]& t7 w0 B- L1 A% R0 V2 ?
so:
; l- f* s! w; |* ZbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s$ ~# J, ^+ _! M5 G1 ^2 D) Y, l
0 m' }9 s; T9 N, g7 x4 t' g
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx- p: G0 F  d  E; g/ L9 f  f
但这是不正确的8 T; Y4 x) r) K# Z! {9 ?
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- l4 D" ^8 c3 |, |0 i! ]
....' _2 E2 Q! \9 C& ~+ K4 S
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:+ X. H) Q7 U' m1 m; c- R
这样算混淆了微分和导数的概念。1 m7 ]3 {! ~" `8 x* \

$ m" ^( {4 }) ad(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- e# W/ ~: L6 d0 R3 v

& ` a$ z- k% R, B) T1 {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:2 x; Z3 s, K( {* y  k
这样算混淆了微分和导数的概念。' y* ^' T& ]1 D$ ~, H* G& [; k3 x2 h1 D
5 i4 }9 }6 m8 X+ r
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( [  Z6 P! R2 i0 L, l$ g4 f! c

/ E0 H  {$ m7 WAh, you made a mistake: it should be:
, N$ C0 Q' O  N; y
& F# o6 O4 |3 d# n: {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:
. ~. l$ y( w9 j: d, y* bYes, there should not be " ' " after "f" and "g", it is a typo. :D
N4 X7 I# e; T `
) K1 _( `) }" [% Y VBut 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:
0 c% f) Q& K/ R2 [* r8 f# P8 m
3 m" z5 d) H, M! t% v0 q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。( h" ^3 }8 m* k5 Q
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 23:24:6 p% c- w7 s/ s; C% N& f6 n
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。

账号		自动登录	找回密码
密码			注册

数学高手请进，多谢！

浏览过的版块