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:
9 S; ?( `' H3 F, C( y7 K4 ]请教大家一道微分题，多谢！
2 Y5 q5 N3 ?, n; `
% `& ^% I- I* F# A9 @d [(a+bx) ] /dx = -kc +s
M/ Z2 u. B6 z! D7 Hwhere: only x and c are unknown, others are all known, requiire c = ?
3 U+ D3 e+ y( ^/ w5 N; G+ [6 {
/ A! P; F. p0 T3 E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:- q% M* @* b# m0 A2 w3 ]1 F
it is too easy:! ~# _& D% Y( P1 k7 M3 T

~# G% H) L) f) b# ed(a+bx)/dx = b) A' L: t) W4 \. m. N- a1 k) q
- Q6 R/ d& A1 V4 z( m; x
so
1 a9 e& ]9 {3 _* V
: s# x9 e: h9 S$ U! p3 f0 a8 Zb=-kc+s8 B8 P( ^6 N" g3 N! u( }
2 x+ F9 M. Y. H: n' A6 _/ B
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: m5 W# z; D/ Y! k: f; G' E
请教大家一道微分题，多谢！
0 [, l0 K" M0 O1 x$ |% z9 ?2 j) p
) ~+ r& @; _( M5 H! H0 U$ bd [(a+bx) ] /dx = -kc +s
+ P6 ]# A, m4 K# Z7 t" U) `where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
; @/ e9 q; n; J
3 l/ H; j& d) q7 |/ O% M多谢了！$ F& M' r1 Q$ v' J9 `3 w

) m2 u( A1 c% E/ w) W[ 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:
- M& ?2 ]: g& v4 e; _% i* ]sorry, did you post another question? your statement is still not clear./ s3 M4 O; G& H: L
+ `7 [9 ^% \( H9 ~3 y
a, b, s, k are constants? c means c(x) ?/ e2 t# O" J) b# F
4 u( d s7 ~$ ]. p# w3 K
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:
7 g# \! s. }# `, z' p( a对不起，写错了，忘了变量c,应该是
/ \+ Y& G0 X0 ~d [(a+bx) c] /dx = -kc +s ; x" o4 a% h+ d" C

/ ^2 d: G; Y% d8 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:1 n" i% E: j! a/ a( Y3 C' c! b
do you mean it is:
+ G% i% X1 a( ?; I( C' U1 v3 Y$ A" M2 J
d { (a+bx) C(x)}/dx = -k C(x) + s
R, \8 \/ d4 p/ z- S1 S' b, B
" O9 G! {- `0 O- b& xwhere a, b, s are constants, you want to know what is C(x) ?
. |) W* E! [/ B1 l
. s1 K5 C6 c- Q9 l7 B9 Dif 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 14:33:5 }( H+ |8 P/ R% s: Z
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:
# D) B# R# y V6 X0 |* g9楼的答案有点问题吧。5 [. H4 m* _/ r0 J. F$ K/ v
您说：( Q' T8 @3 F+ {" Q8 |$ S

# L: T! E6 Z* d8 td{(a+bx)*C(x)}/dx =-k C(x) + s* ~# {$ d, g8 e8 F& S
so:
" a, ^: F" q1 Z0 V" mbC(x) + (a+bx) dC(x)/dx = -kC(x) +s- b6 o2 p* i. a
! ]6 D+ n: [; q. I- b$ n
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
) M' i0 [, _3 P- h但这是不正确的
1 q7 ^1 E1 N' V4 U9 s6 vd{(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$ ~( T5 P/ a9 u& i# y* U4 M
....
. s! y9 n$ P) o3 y0 D E! 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:" A4 w' `1 r- a
这样算混淆了微分和导数的概念。
9 e4 Y H* {4 o2 D7 o2 X( v/ a$ O' z; j/ D/ I$ B% b: n
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" S; g0 r0 N8 R

# w- N/ z [5 q) p/ w% ^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:, q, m" d4 r; t2 T( t, y
这样算混淆了微分和导数的概念。
) G4 @7 U$ I; O& K p' g( o' }5 ]0 t
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% O& f9 C. x" q: n

) t$ L( c: t3 hAh, you made a mistake: it should be:' l7 p: @ M1 F
; a# V% c/ p/ T4 e. s
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:* h& Y, r# T$ F: M- n8 g
Yes, there should not be " ' " after "f" and "g", it is a typo. :D2 R' z; }# B% O8 \- n/ e

# M. P$ M. D1 ^+ `1 p( aBut 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:6 V% T9 D6 ^4 o! V

2 @1 A* N, t2 k- i0 m+ z+ [8 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。
- T O7 }- X/ A4 ^) @0 ?/ N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:
/ F7 q$ f' Y6 W# Q% ] E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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