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:
' a, u% l2 g8 \* g2 i请教大家一道微分题，多谢！! x5 c% @, u, N9 c2 Q" I

8 K$ a' c7 O% k6 id [(a+bx) ] /dx = -kc +s ' Y4 @4 r* M r. P* R6 t% d; N
where: only x and c are unknown, others are all known, requiire c = ?
' W; I% M1 C% L5 v8 T
' L& \) i0 e5 @& X% \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:
, v* f# Q+ n- `% i$ S0 Cit is too easy:. M) D4 a! g+ k) @% L) C4 [8 l, M
2 } ^+ w7 O# h2 n) P( J5 e
d(a+bx)/dx = b; ]' A' H7 H3 B+ h
% C% l7 m7 S! f! f3 j' B
so; d( l- G; B8 U6 T3 ]
3 V) E' N, _: R$ X# [; D: r
b=-kc+s0 O" U2 |2 [; q& [0 y" o1 o

|& L [' d9 K* {' k% u& pfinally: 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 20:41:8 `8 ?" p3 d4 }' j0 \0 c
请教大家一道微分题，多谢！
2 B+ v0 q O' V9 k9 V, B
7 V/ l! I( l7 B. g* m4 Bd [(a+bx) ] /dx = -kc +s
. ~9 V0 {$ O, G" X* ywhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?. f& d7 P x8 Y" ~

1 B9 l6 |0 s8 A6 i3 i$ Q8 ?多谢了！
# A8 q$ t" Z/ h( A# U( S0 G( S
* q9 D8 T2 M: I+ y( Y- S[ 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:5 }3 B9 B' F! X! g6 W% s
sorry, did you post another question? your statement is still not clear.
) Z% c0 s" M. Q/ S/ n6 R& P: @1 ~; h8 }; h
a, b, s, k are constants? c means c(x) ?3 k' x6 y9 H' s" ?6 j5 o% C5 [
7 V' }0 t7 i1 F' s$ M. D% p
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:+ B7 }8 Z8 Y+ `; F+ @
对不起，写错了，忘了变量c,应该是2 t; x* ?# b( U2 A' k
d [(a+bx) c] /dx = -kc +s
/ k" |& j' n8 I6 L
+ W0 x# f/ I5 J( I, Q/ _7 F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 B0 l; \( [& E( r/ W5 t8 bdo you mean it is:4 N5 T3 |9 @8 d$ P
" s& b8 a$ G: s6 M$ G% e7 h
d { (a+bx) C(x)}/dx = -k C(x) + s7 F" X9 b. D s! V7 \7 c

9 e& U' P$ a7 V0 F3 N$ Hwhere a, b, s are constants, you want to know what is C(x) ?
3 o) U/ J2 M3 r/ @
5 [* ^ [+ F0 }; c/ ]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 14:33:6 Q# s$ H6 Y1 O8 f7 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:  ^7 p% o, U* B( S
9楼的答案有点问题吧。# k: c% J: A0 @
您说：8 N; ~6 s0 P9 g3 u. G* B- S

" ?/ |! L7 m9 ]+ G2 _d{(a+bx)*C(x)}/dx =-k C(x) + s
  i% a$ I: V; fso:3 `, s  Y6 e( g2 C. b- P
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
+ i) ?. b& x7 Y% I9 f8 F  [6 M/ ~
9 q( j1 F% ~  n1 [也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx9 B' N9 A0 t8 H
但这是不正确的! ?8 L' j) q6 a2 f
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8 g; b! B9 g8 U3 v
....( P0 [% A/ A5 @2 `1 o# v* a# \7 t3 u8 n
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:
+ ^3 C' [. j" j% }7 A这样算混淆了微分和导数的概念。 x! Y# g6 k5 m
9 x4 y3 Z( t% u0 Z4 S' `& 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
( f' v9 O5 d. G' ^' U
) D; b, y* y+ t: T1 IIf 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:) }" W' I+ g9 b3 ~3 r
这样算混淆了微分和导数的概念。/ r. V4 ?& d: Y

6 v( d, n# U7 j9 }8 z8 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
! m' u8 D1 a7 ~- i, ^
5 R' [' c+ i' m& X' t3 jAh, you made a mistake: it should be:
' _) Q# M- Z5 {& Y$ f1 I1 H/ w8 _, d2 }4 s0 y8 Z
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:$ e2 a0 z @( r' w8 N; }7 o
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
: O5 S! _5 G4 `
9 k$ y2 K1 N* w3 Z# \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:( B2 M) Q, e& d% }
6 Y6 }3 N9 ^2 l' X' [. R
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。' K- y6 Q l+ N k* Y4 r* E
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:
' Y8 ~* P" g; X2 n4 z' L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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