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:
# {# o3 J& e+ ]' s+ |: y请教大家一道微分题，多谢！+ }% T! V% k8 N7 R( _0 }( h) n) ]
( S# n& B9 X) L4 s f
d [(a+bx) ] /dx = -kc +s - r$ @/ E% K* x/ S4 T, s
where: only x and c are unknown, others are all known, requiire c = ?

; p% @6 q" ?4 Y- 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:
# @% t( Z! h+ C. b: eit is too easy:
4 b+ U8 l  y$ s# B& R
' n3 Z' M6 S9 p6 p: T( \d(a+bx)/dx = b) z! V- A2 q0 [. ?& S
/ t4 X. a! K/ K- N, K1 M
so: l- i+ O: V1 f9 ~" w8 {% M) n
6 c8 g( V# g3 O. [* M: I# N
b=-kc+s) e  [6 c, _+ M+ @5 n  a. i

0 }3 a2 k6 f; I& v) Tfinally:  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:
) `* ]; W3 C. B  ]$ b, @' H7 |请教大家一道微分题，多谢！
( J. l# J2 m7 O9 }0 h, m' i/ T% W* L* M% o- s! k' s* S
d [(a+bx) ] /dx = -kc +s ( `) n3 M" h% P
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
6 M: u4 f& l0 h; O* ^  L" K$ N! |8 s$ `8 r: l
多谢了！
: w, w; U2 A! W: m4 @7 G8 @" G1 i2 n2 M1 b8 w: R
[ 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:7 A- O+ A+ {! i0 x6 Q
sorry, did you post another question? your statement is still not clear.: k" v, G: k, R8 X) F9 t4 Q

$ U# W7 s4 n) c/ Y, ca, b, s, k are constants? c means c(x) ?. x) W o V0 i1 S! q( L

* ^& o5 V% M* G, O' W+ ]/ W8 w A$ ]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:
* |, g/ P, X8 y9 \" q: L对不起，写错了，忘了变量c,应该是5 M" e! _* g0 O, O( a" `
d [(a+bx) c] /dx = -kc +s
2 m0 t. M: o" b
) w8 c) |3 |; H4 O0 f( ?" o9 r$ o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:
0 ]3 D9 M+ q- I* U) U" H# a/ udo you mean it is:9 q. Y) x' j8 b0 y1 K* v

. ^" m9 K$ O! u2 fd { (a+bx) C(x)}/dx = -k C(x) + s
* \6 E: Y1 B8 }) a, C
6 |9 {) R' Z: u, S/ Ywhere a, b, s are constants, you want to know what is C(x) ?3 r+ m6 g* J! T- J/ \1 ^9 z
8 n4 w+ V* c* q5 e# \- [& N
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 14:33:0 A* h; n! L2 h( e0 K; _. 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:
- _( f# a" O' p$ q6 m* v9楼的答案有点问题吧。
# t; @& {6 Q: i, N* Z6 ]6 P7 e您说：# t2 \# A7 m; Y) w. L; o

/ E5 t( e0 a3 ?, H( f2 K6 ld{(a+bx)*C(x)}/dx =-k C(x) + s
, @' c# H) t8 S! J9 p' lso:
l/ l4 l* B( I/ ? pbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
2 z: D/ \* r" P/ t) P% c$ E3 b7 Y5 O
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
$ S: D: o g2 f) x但这是不正确的
1 Q# ?5 p- K! ~5 M) O9 ~- [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2 ^2 [2 ^( T9 a- X
....
1 p# P7 | T- X) R7 Q4 Rd{(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:
5 z+ Z ^2 Q3 }: w0 Z1 ?这样算混淆了微分和导数的概念。% k0 m. v( q% n: h, v1 ?" T/ B

" U: Q$ S, ^$ d! |- w0 _# Kd(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
7 l/ N; p4 |2 C+ p7 Y: ] u& s% K$ B6 F. Y: Y+ M
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:$ s5 U& ?$ y y) S! E
这样算混淆了微分和导数的概念。
; J1 E8 h5 _5 I% N, H5 N* G
: Q# q4 U9 s6 ^$ T# ed(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 b" u9 ~5 R% p2 M' N$ y( S! g4 [2 m3 e; z
Ah, you made a mistake: it should be:" y) r' G1 {- r$ T5 X; S/ s- X0 X9 O

( f0 z! q. |! b. [5 d4 J4 Sd(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:. J+ P5 {& g1 \2 a9 r
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
. j( M- b F! O% H4 ]0 G
2 W5 ~8 [+ B. i; P2 KBut 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% A; G- S' r! K% a9 X2 ?
# N V5 h, G; U& v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。
& I2 d( B( }5 b2 a, ]5 V9 j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:
5 H' S" s$ C1 {; C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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