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:
, T: w/ B; |3 o4 F% k请教大家一道微分题，多谢！ u, w1 I$ t# `, l' O6 c) X
; Q% b; i# {( m$ @; n; S6 B" f
d [(a+bx) ] /dx = -kc +s 7 Z: {* s2 O/ x+ M7 n# [- Y
where: only x and c are unknown, others are all known, requiire c = ?
" P' R" U% S& w* a5 f/ \/ W/ }2 J& v) b8 s5 u( P0 Q$ _& u
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:2 }! r0 P( `, X+ x( N( z
it is too easy:  i$ c) {# R0 V

) Y) r9 ~7 P% _! R3 [d(a+bx)/dx = b3 b; s/ y( k  }- `6 `% h
" G8 Q) D- d3 W- ^
so0 [7 d5 E2 P; }; T
$ V2 z! ?1 P( k! ^- U
b=-kc+s0 T& J9 a0 G2 i% ?
) `6 h+ H. I! ~0 a7 W
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:
: g6 A: B/ q+ \+ `请教大家一道微分题，多谢！+ U9 C9 v6 Y. ~# W3 }3 p

4 U, K) O* j7 H( |d [(a+bx) ] /dx = -kc +s
+ P* v8 T6 b8 f4 c6 D9 p! Xwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
' f" R9 X' Y- e6 j8 }! Q1 B1 q; ?4 r; N5 C6 l8 f7 |! I
多谢了！
4 A6 O& ~2 ?2 h% |0 p* F" z
$ l# i9 d& G9 M7 |8 ^# v[ 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:
1 \" \4 _. |) _$ o7 h( Xsorry, did you post another question? your statement is still not clear.
% }; O3 Y% ~9 y; ]9 b* [; ?# }" T, h- j/ }* j( @3 m
a, b, s, k are constants? c means c(x) ?" g8 i& Z* V& K& M* B, Y
$ T, n$ S" z) b) X. 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:
: i6 Q2 ?4 M. Y& h3 j对不起，写错了，忘了变量c,应该是: f6 _$ {4 F* x6 V
d [(a+bx) c] /dx = -kc +s 9 g1 R& X! L( d* x" D5 S( n7 b; A
9 ~0 W' d( D1 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:
. w( z8 h' T' t( pdo you mean it is:
# m' p8 P, b7 g2 M6 C0 g/ z5 [" N
" Q1 D7 Y' |* [# h' p. ud { (a+bx) C(x)}/dx = -k C(x) + s
, D3 W- y! k A
# e. `8 o; M! m3 J$ Uwhere a, b, s are constants, you want to know what is C(x) ?
- a1 x1 y# v# t+ @- v4 T5 m9 a0 G* @3 U0 N" o5 T/ t P
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:
( I2 Z M* |0 {2 x s* 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:
9 P& Z  a4 H, z. L5 Q9楼的答案有点问题吧。
- ]; t& {2 A, C$ U2 P7 e您说：8 h. R* E: [& h9 ]6 P9 i5 @% R
$ {* m7 a! k$ x  `% [( h9 H6 J/ B
d{(a+bx)*C(x)}/dx =-k C(x) + s
& h* |1 S9 O4 m- ?so:- Z% p* d8 J7 ?8 M! z+ R
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s( ~2 ?+ }8 i0 M  |1 m; k
0 z& X& f$ V- a- }4 f8 G2 q
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx0 F9 N  v2 u7 V  X, X" ]; \
但这是不正确的7 F' W* J: z, X- v9 q- i9 Q3 Y
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% O j" r! e$ k. D C
....
3 l+ `! P) |4 F5 i# ^+ gd{(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:
$ l& j0 d" {  H2 ]8 x$ W1 Z" ^, W6 \这样算混淆了微分和导数的概念。# c; }/ }- ^* ?) f. T* }
( P4 r& g  x/ L3 p
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/ o2 E; N  K3 }# |0 ?  Z# w! O
( R" @' K9 y' l# }- C" u
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:* u( n" T! A- M, G
这样算混淆了微分和导数的概念。0 k# S$ b( L' {* R4 ]

& f4 q( x) s$ C; @: L; z% ]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

5 h9 P# V, @9 {8 o& e3 ^Ah, you made a mistake: it should be:
7 f' U/ D: p9 Q1 w+ U% K3 T6 v6 ?) ]0 ^
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:
3 j7 ~! I) u7 ~5 A6 M- w) EYes, there should not be " ' " after "f" and "g", it is a typo. :D j# u% Y2 h8 c
, d% p6 m5 s9 r- Y0 M* J
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:$ b4 f2 @3 }& ~+ G: G" z! @8 P" y

, W3 c" a+ f) A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。# n* W7 |, Y* @, 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:. O7 X6 G7 Z9 C% E6 M+ w
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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