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:
8 w/ D' L7 c, H5 L1 c( G' R* \请教大家一道微分题，多谢！
5 i. w  q- ~" C1 e, G+ H3 H6 s% d, ]0 x3 s
d [(a+bx) ] /dx = -kc +s $ \# ^; d' B& g8 a; i1 o
where: only x and c are unknown, others are all known,  requiire c = ?
  V* E0 \8 w8 P) |: s5 R& |
7 o( `! z9 D7 D; I7 y8 l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:) V0 |, k# h. ]$ F# r
it is too easy:
: W! a0 E7 W. N1 [4 t0 o
( ~9 e+ E  o  F8 z$ rd(a+bx)/dx = b
0 k! Q, {$ l1 _/ `7 J0 E
! S6 B2 ?& t3 j, fso
4 \* Z. I2 |$ H% A8 R' G  A8 B0 ^& W( y3 v
b=-kc+s
1 D7 l- X+ O3 }- f
, n7 S  r! S0 F4 Z3 v, Qfinally:  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:* Q# x& f% v' u) R
请教大家一道微分题，多谢！+ ^6 ?, m+ @" c; l$ Z5 M5 l4 u7 E
# r5 O% I: a f& x) K
d [(a+bx) ] /dx = -kc +s
8 z2 S0 M1 B" T2 e+ ]2 G8 Uwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?) Z: m" _* [$ f8 Y$ c
9 i' v' J& ~6 o# {! b! r1 U
多谢了！
8 S- j8 Y* m* E) Q. @' |$ h% i0 B9 i& C' _8 k
[ 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:2 O  Y$ u. l# }+ L' @. e
sorry, did you post another question? your statement is still not clear.
9 L3 o0 r* e; B' w' o; g
0 j! [  `3 U! @" p6 Aa, b, s, k are  constants? c means c(x) ?
* I  [9 V! U; e5 Z$ M
' U  Z2 q! N" S$ b$ o+ aplease 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 ?4 Z; T$ m. \5 V
对不起，写错了，忘了变量c,应该是1 N% Q4 T: P. o# z8 w
d [(a+bx) c] /dx = -kc +s
* E# K* a1 @5 W& F. @0 N
1 b6 R* M% D6 C) i* D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 F8 r8 w4 E- hdo you mean it is:+ ~) p3 o4 w7 n: z3 y
. W/ Y3 H5 o; q# _) U7 K  A2 ~
d { (a+bx) C(x)}/dx = -k C(x) + s
+ K- r6 p1 p  M( J2 I5 k: H. t  o# ~7 F7 q
where a, b, s are constants, you want to know what is C(x) ?, p  N& y2 _  V6 R+ Y4 c) {! j5 u2 n: R

/ {7 H2 ?" i- x0 I: J8 {( H. mif 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 14:33:9 j! h [/ f/ w0 g# x
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 @& w0 y2 g: j" k8 D) T: a
9楼的答案有点问题吧。- Q' W' ?, `5 ]( }
您说：
) D- x6 j( P: q# ^9 ]" \9 L8 `/ Z
' o. u7 g' z5 Yd{(a+bx)*C(x)}/dx =-k C(x) + s
0 x* @* q; r% x$ Sso:
8 f( m: B6 ^( J0 U4 Y% M0 HbC(x) + (a+bx) dC(x)/dx = -kC(x) +s0 N2 Q: r' q, @& \# @8 s
, H* t' }/ [% P9 F
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
' x& q+ p: M( b/ @6 C! |但这是不正确的
& A: q; ^# t! `/ {- C3 z4 k/ Kd{(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% f; w% f1 K2 }! E. T2 O
....
5 o/ [5 u; k3 ld{(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:$ _, H1 P7 [; Y* J( g6 Q1 K/ |
这样算混淆了微分和导数的概念。
3 [9 c7 s4 p( ^1 ]- T1 ~+ A. f
" E! k6 n r8 {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1 w1 N# l2 v( g3 g' V" M: g
+ T; G# | F. n( q) i
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:
3 f3 C: K3 J1 J& _这样算混淆了微分和导数的概念。
- d* B. U x/ u
# M, e; g9 G/ w6 ^( Jd(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6 @6 |# [+ j# n1 `. m0 K+ h9 d
# z1 L: S/ h( d& R0 q
Ah, you made a mistake: it should be:2 Q% U" F& z) g! _2 i% i- Q
5 c" Z w2 h" U0 Z9 k
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:% b8 ~% H F5 k. M9 a
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
$ b* K3 e2 i; y6 Z+ P+ G/ l5 E2 E, O( b
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:
0 b& p' a* k L6 }5 u' G0 F' ?1 q5 G4 ^6 i7 \1 G: T( D8 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。
8 n0 A7 [* Y# h" t P4 D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:% t* {1 L* [7 @5 F
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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