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 20:41:6 Q3 V" F$ }0 P+ T; q
请教大家一道微分题，多谢！
1 S: x0 {& _- i& g5 B" L) Q5 c5 k0 S
d [(a+bx) ] /dx = -kc +s
1 Z# t7 x3 g( G2 [where: only x and c are unknown, others are all known, requiire c = ?
p; W A$ r! X6 [0 x' b6 N3 H4 Z1 r; x; n+ n6 `# M5 z
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:5 S$ ^5 o" u* [1 S" M( |4 {
it is too easy:3 m9 ~: v& A0 ~0 m2 E; z
5 X) d& I- }+ ~4 \/ c( W5 w! @1 ]5 V
d(a+bx)/dx = b
O' O. N" Y4 O/ Z& K% r. N- Z: |, | D$ f7 |
so! i9 }* h. R7 J! ]7 c- K. s

( }- O& n) ^+ j% Ib=-kc+s
4 N0 X! \* ~$ u( M) }
* M9 Y) _) u: Y6 Ifinally: 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:( i( H# n. s/ L  J, J: z% R
请教大家一道微分题，多谢！
/ E; U% f& p! ?) T% M
; A4 R; `& S' v8 W* Wd [(a+bx) ] /dx = -kc +s 8 u; G; _! i, Y( ]
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
; _+ K% F3 W8 n, ^6 t( D6 z4 Y- X0 x$ L$ a# R" r  A
多谢了！
5 s2 J, _" h% y) |+ T3 m5 ]: i1 c) n: C
[ 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:, N$ ?3 y a5 A- C: u. P0 a% t
sorry, did you post another question? your statement is still not clear.
) G1 f% X8 t- f' T) A
# r. c( m! H' L$ Z# |9 La, b, s, k are constants? c means c(x) ?
5 I" b- X% ?1 X: c8 A9 R& j" v1 N
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:
1 W5 o0 W a' r5 L. Q9 c! B对不起，写错了，忘了变量c,应该是2 O5 f% }* g) Q, f6 V) N
d [(a+bx) c] /dx = -kc +s / v8 Z( l+ m; ?( g6 c

) j: v+ L6 U- Y. C# `5 T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:2 E: _; V M1 P. L0 Z+ H/ S+ k
do you mean it is:# Q: {/ h6 A6 x3 y4 ?
. `. a( V6 x! k1 F( ?! ~
d { (a+bx) C(x)}/dx = -k C(x) + s) \& G1 s6 f. }$ Z9 r# c
8 e1 p9 }* v* p7 d$ k
where a, b, s are constants, you want to know what is C(x) ?
- T5 N" p5 l9 \- L8 [
- J; x1 }% _, P2 d) Zif 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:
3 {) V) W4 v) V5 ^& o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:
+ L( G, h- r& y3 Z+ U: f- r9楼的答案有点问题吧。
3 s8 [, q' r5 [( N8 y, j您说：! F7 O/ I5 ~6 L6 M8 C1 b

( \  o3 V8 h& Y3 e0 o* ^& yd{(a+bx)*C(x)}/dx =-k C(x) + s
3 L( G' P- m- A) G; ^so:
0 O4 ?! o- d9 J" s7 Y, @bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s# U, V- u$ g( U! b0 ^* C
3 e2 }9 C: S2 \8 ]" a: r0 f5 [
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
) a. {1 I6 J' N  v2 T; n但这是不正确的
) R" U% }- A- Z& \2 Od{(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0 k6 I6 Q/ o6 s
....$ ]% c# M+ o1 |! d. A: R
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:
- @' J0 q- B& x' h/ X) I4 o2 o这样算混淆了微分和导数的概念。" w8 A5 Q; U* V0 b
7 F8 e/ C" b5 H" q- i. d+ 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" E7 \; y/ A! M) ~1 b, r6 L: e

: C6 s, ^- ~) dIf 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:6 G2 r: w( D# Q3 p, Z* X
这样算混淆了微分和导数的概念。
! o6 j! W2 ~+ i# T
' |7 L5 t% x. Zd(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4 Z7 J' G5 {9 r5 y2 n$ Y! k" p4 T, n
# G7 M/ c6 m' f0 l; s, }
Ah, you made a mistake: it should be:
8 B; F, g5 a* M5 M4 ], O0 |# {- T- G- k; B
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:
. B4 ]' A8 @4 ~; FYes, there should not be " ' " after "f" and "g", it is a typo. :D9 C* h/ T! `) N6 R" ^& A

0 |3 x" X8 {0 @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:$ N1 ?: _3 Q' h

: M+ r* K+ g4 z5 P1 {- K0 L. s+ 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。! f( r. A" A& A+ x0 N/ I$ M
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:) |$ Q4 W0 e, I. j& P. Z
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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