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:
) T6 b1 Z, N1 }$ Y请教大家一道微分题，多谢！: `: k" C* _" \1 ?: E6 @

8 |4 Y. X8 Y/ b% L7 r5 z- td [(a+bx) ] /dx = -kc +s 0 n/ Q) \8 K. A6 W) K' O
where: only x and c are unknown, others are all known, requiire c = ?/ O3 _( z& ]+ M* ]+ t8 ?
% t1 a/ E9 X: H+ z3 x" H
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:0 a# o* _- U7 E' J& h
it is too easy:4 @2 x; a  g1 Y& N

* Z5 ^) o% C- Td(a+bx)/dx = b2 y0 u# g# T& M, _4 H( n& k
$ O  l8 g: ~6 ]; ^4 X& s7 {
so. N& j" Z& {  T8 x' m; v2 ]& {

6 x' u6 k7 E4 L4 ~b=-kc+s' }0 m9 {$ e# x

- \( I3 e! u% K  d% H' g6 ^5 Vfinally:  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 f* _, ?) D6 y0 ]请教大家一道微分题，多谢！# H% A' N% |* v' z

- q3 G% p9 s* J% ^. O# g* kd [(a+bx) ] /dx = -kc +s
9 v* `1 d5 Z" z5 Gwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
3 }% d, \, j, i! J1 r
; I" x6 p) E M( Z) E! W多谢了！
7 ^4 m% \$ j- h3 r# Y9 R& G. A! ^! G
[ 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:
$ p4 G2 R) y: W6 Dsorry, did you post another question? your statement is still not clear.
. U6 J ?# ~2 y; {1 E- |2 \& P1 F- w& H& ^' L
a, b, s, k are constants? c means c(x) ?
! C, B" |( n& g) t
/ q' X7 s# A4 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:
" f9 c0 _+ ?1 b7 ^6 s对不起，写错了，忘了变量c,应该是" }# ~, B' v: _% t
d [(a+bx) c] /dx = -kc +s W" a! M# d% p: V3 S

r, P' a3 k6 j" G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:' g* g8 R' |, L' P; A% k8 s- H* G7 Y
do you mean it is:
) r/ N: P% b7 E8 I4 U% u+ y* d/ P* O- z) m- T
d { (a+bx) C(x)}/dx = -k C(x) + s& c3 P2 ?* W# s7 Z2 [/ y! J

% |: V4 B" d# |3 k! _/ w9 {where a, b, s are constants, you want to know what is C(x) ?# y, g3 s4 ^1 E9 X9 Z$ U( }: [
' D, n! u$ K' L
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:
& i! u, Q' ]) i3 X9 M% H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:
9 u8 |/ s$ Z) m9楼的答案有点问题吧。0 |( ~* ~1 i+ [7 X5 B
您说：9 d, G+ R m; ]% [, D
4 V. G3 N; e# {4 m+ `$ t
d{(a+bx)*C(x)}/dx =-k C(x) + s
3 y7 b0 R. J. {/ ?3 M- Z! Pso:
, ^8 Q f3 P! |$ K! z( b& u+ YbC(x) + (a+bx) dC(x)/dx = -kC(x) +s! S) F3 n$ _8 r" Z* I

8 D2 ^. ^* S' M& ~也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
/ P5 y3 d: ]' D* d/ x/ y3 n但这是不正确的
9 ]# X$ r7 N/ M7 b( F9 D" ud{(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: i+ N, {: Y9 R1 W
....- J" P6 d$ l; y( A. L
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:
0 X4 c/ J2 z. e" p这样算混淆了微分和导数的概念。* m$ N% V5 k. G8 Z- L
( c- O$ a) L1 G# Q
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
- O8 _% W0 x* ?/ k' v- ]* C8 \" F+ v9 B" T* |
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:
; {! }. P9 u1 y) n9 v1 f0 |+ Y* U这样算混淆了微分和导数的概念。
5 x+ `! E* M. g" a4 O
, R, u7 ]+ q' T8 ?* H. W- r5 n3 Ld(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 F' H- l" Y5 U. ^: g5 o; L
8 z6 [" p, z3 E) LAh, you made a mistake: it should be:& K7 o+ C0 ^. _( d

1 }# }% V4 W0 T: ~& r5 h6 i- 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 20:51:2 w, g, W) y7 Y8 C4 i9 Z: W/ J
Yes, there should not be " ' " after "f" and "g", it is a typo. :D# u/ |7 q/ m% R& M; h4 P1 s+ W
. c$ B9 H! D3 H
But 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:
! }- M* u; \/ ]. ~+ D$ w$ I# S/ |* L: t. F2 ^9 l3 G
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。
) L2 Z% r. b' z( s& N% _5 P. t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 Y8 u- B# c2 l X4 Z' ]7 ^+ q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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