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:/ k0 x8 D% z' }5 m$ N
请教大家一道微分题，多谢！
1 n& G- F6 q R! G& x$ A. g8 E! U
d [(a+bx) ] /dx = -kc +s
4 G' @) r- w- G2 uwhere: only x and c are unknown, others are all known, requiire c = ?
3 X9 V4 k% x! G( R8 A& m+ J
4 M% Q4 b% Z, F/ c. a3 ~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:
7 A+ k! m0 E6 y8 Qit is too easy:
" U- C, W2 c* R/ S! _8 I
" v9 S' E# a9 ^" v; {1 {d(a+bx)/dx = b
- N/ \- f8 n8 Z+ @, y% E' ?" i; s* U: _1 ^: B- [
so
. j- |/ h; _5 |, ^7 X7 H. l" b5 V7 s5 g; @7 A( p7 F/ q8 ]
b=-kc+s
' v) h% C, W2 O) _. N2 L( T6 ]. L4 }, W" C& c
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:# S, a1 T. F. n# x& S* Y
请教大家一道微分题，多谢！
) p% b( O! C1 {" w" ]! C# q! b5 k2 u1 t v: s* X
d [(a+bx) ] /dx = -kc +s
9 h3 t8 W; r' h; g3 ]* |* Awhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?7 N2 u+ b1 k6 a# j' e9 w

# ]8 Z" b; ]6 x% R* {- `多谢了！9 q0 E6 O- ~% E0 A/ ^& f! u5 g
; [8 \+ Q: B6 O* [4 U+ i. i) R
[ 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:
/ R8 _% F* c! d( Z% Dsorry, did you post another question? your statement is still not clear.
, \& |: v8 w1 M$ f; s7 g' F; s3 o, B
% |1 `: ?# O! S1 b: l: Na, b, s, k are constants? c means c(x) ?; d: p, [) Z t8 O E& G' R1 P; l
( N7 U' b5 Z# y1 D
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 12:30:0 l! U% l* f% W) z
对不起，写错了，忘了变量c,应该是
1 n8 p: D7 ~0 {# H, I) Id [(a+bx) c] /dx = -kc +s 1 r* q3 y* z, @1 p6 v* O$ Y

~: Z8 B" v/ {& E$ 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:
& g q! Q8 e. C" ?do you mean it is:
! y8 D/ c* k; Y' k0 `, |
% _& a# S: F: Q) qd { (a+bx) C(x)}/dx = -k C(x) + s
/ g, e- ^# A+ p# ^9 r9 ~0 y5 }( T2 ?5 `: n2 h4 o% S; ]7 c7 t0 g
where a, b, s are constants, you want to know what is C(x) ?" L) M+ Z( T% ^4 j+ K
. C# J) I1 Y9 S$ S& H; _
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:
$ W, ]0 E( z4 @& L& G0 v$ y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:
( @" m/ `6 |7 b: u0 g2 _9楼的答案有点问题吧。
' z& ~* n6 L) x( H# [5 R+ e9 i您说：0 x6 F+ l5 N( o& D% A/ C
: ~) D: z: h% z
d{(a+bx)*C(x)}/dx =-k C(x) + s
  x% i6 X" \" R2 q1 X& G" @' R) Aso:& v. u  z) K" `6 }4 |$ x
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
. x! q- B( k# E9 V- `
9 S4 w2 l6 p: q2 Q0 l  C6 K, z也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx; y8 }4 a6 p$ u, O, V' B' M
但这是不正确的
: h/ p! E1 {1 g0 B  C2 l, o8 V# Rd{(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
* Q' Z- E7 w1 @) Q4 a$ A....; b2 m ?% V" y6 C+ c5 V- g
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:. e: y; f+ ~' u/ f* a3 ?
这样算混淆了微分和导数的概念。
: m- |4 c% o" N+ ?0 e7 p# v# y" t7 `" _% H$ y! N8 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* A$ B! ^/ H5 u4 J e' z

# B' [' s' W( }* DIf 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:
1 c$ @9 y/ h8 d/ f% k这样算混淆了微分和导数的概念。. B5 t6 y" Q5 t

: O( W4 b# _8 [# p# fd(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 ]- s/ l- q4 q2 w0 H$ m
- D+ Y2 ^; D1 {
Ah, you made a mistake: it should be: w% Q) w: F e. N1 P

8 `- M( b4 q0 @$ |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:
) V, J& k5 `# F. \Yes, there should not be " ' " after "f" and "g", it is a typo. :D
8 L+ |* q: y7 U' A8 t
4 V0 h" J: |3 ~& |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:; q2 J% _( } r+ ] _. \

3 g" m2 t) f7 \, j3 E- I+ n' o. H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 g" h: L$ g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:% r, `7 M$ m" v5 \: n
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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