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:/ e7 @# H9 j) R: F$ ]
请教大家一道微分题，多谢！9 m0 m% [ _* D" K6 [( B
) }" [, C- _7 m4 t- ^2 E5 X4 c
d [(a+bx) ] /dx = -kc +s ; E% w1 t/ l* W" f' P1 T
where: only x and c are unknown, others are all known, requiire c = ?: Y' @ S1 g' Z( U

$ x7 _3 ^/ R, q1 ?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; y) V! v- u* H* P5 C% h  |it is too easy:" k  v' y2 j( O4 Y: v

) q; M3 r" {' h8 ~$ O+ ]. Y2 ]d(a+bx)/dx = b9 o( b9 D1 t' M3 M, m% T/ q; {

( [7 r) e) @* q7 `, f$ q  uso
4 L- r+ H3 ]$ v3 B' k& @
; Z7 N+ K7 M% X* d# C4 d, q9 b! Bb=-kc+s
7 Q7 a) t8 Z! k+ J5 h# N
- x+ E. [' I! M- K8 T# e% o+ H' efinally:  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:
$ ?3 K+ ]. p0 e$ x请教大家一道微分题，多谢！2 k5 y$ K: [$ G B

+ u# u: w6 ~9 i1 ?2 z$ }d [(a+bx) ] /dx = -kc +s , g1 Y6 ]* l. {+ \7 W9 C
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?) h ^4 x9 P H+ w

多谢了！6 E- V5 d, }5 I9 i8 d1 r5 f
) \! D- l4 E. u; V9 w, M
[ 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:
( Q& _, m5 t: b2 R. Y3 Z8 tsorry, did you post another question? your statement is still not clear.
+ z  P4 G, C+ E5 g' O, w. S2 Y, s7 l  p/ z4 l- M$ j3 Y
a, b, s, k are  constants? c means c(x) ?* |+ l  I8 e) p6 I# F

9 @9 @7 Z  ~9 g( C$ Pplease 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 u% W* k. E8 L# Q0 i) U( V1 k/ |
对不起，写错了，忘了变量c,应该是9 o) X9 o% N7 [3 P7 [( e
d [(a+bx) c] /dx = -kc +s 2 s1 n# G( g5 N/ P

$ E) i! g" J/ ^' X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:* R) S7 C3 S3 L; C! h! L9 g9 t( y
do you mean it is:2 P7 X3 O$ X! [# u- D3 X" [
: c! r2 f3 X- {3 G" C! Q
d { (a+bx) C(x)}/dx = -k C(x) + s
! H- Z( u- s8 b; ~ q
]$ V; u6 w, e( z* ?, Z, Zwhere a, b, s are constants, you want to know what is C(x) ?2 C$ O5 Y; A6 j0 ?+ p" Y7 ^( G

4 m* T4 X3 l8 Gif 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:" L; O9 _* S# ?! ]6 D$ Y7 q
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:
1 z+ B! z. v% J7 c2 b! M- k6 W9楼的答案有点问题吧。7 x& s# u* ~; y( X$ Q3 F! X4 }
您说：
' j8 H; K9 J7 F& J4 Z' z9 i9 H/ ?6 {  F' U5 P
d{(a+bx)*C(x)}/dx =-k C(x) + s
) K. T8 }! v: \3 v( _& B# tso:1 }) y2 F  k7 l  D4 y6 z& F: H
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s  p7 w! c0 R* p/ _( K  m+ f

1 d5 n- O1 u( o3 A也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx5 H) F( B, T, |! w% G
但这是不正确的" ?3 @2 G. k' t6 S
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. P& E9 p& m+ K L1 i
....& v, y! R/ M( m& @0 W. \* D
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:' R. f# u2 a% G/ ^7 I4 q# g
这样算混淆了微分和导数的概念。( C' R- t( e5 f% g( R' d6 ~8 i

- N# b% M: c5 V8 z, H: _- xd(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
& n, Q. |8 p. I7 a% x. e" G, { E' d( `7 l
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& w6 V( o' y. z9 L
这样算混淆了微分和导数的概念。
6 _ P4 Y1 Q' ]3 K# r; S1 u0 v6 l3 O9 C5 |3 e7 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
- U: B: O3 A1 y; I3 { Y' F$ K2 S5 a
Ah, you made a mistake: it should be:
; j2 i0 p" M* k7 `6 x& ^4 i( D) v& d. M2 ]% B* O
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: X' m/ p! O$ C# A) b! [2 Z
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
6 ^1 M5 U* \4 {3 o# N; f# z' F6 D1 i8 q- D; v
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:9 h; G8 V5 V8 D% Y6 E
X' m0 k5 s+ V# ~( Z% |$ T
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。4 v4 `/ g+ a' k, J* P7 P
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:) `! b! w8 K _) d
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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