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:
. K1 l  E- E2 X请教大家一道微分题，多谢！) ~7 u2 |7 t8 \4 w; r7 f) g4 M8 i4 f* f

# t  X' ?7 x( q5 J6 Pd [(a+bx) ] /dx = -kc +s
; J. Y: v+ Z! Z3 f3 G0 Qwhere: only x and c are unknown, others are all known,  requiire c = ?1 F. J) ]& Z1 D+ e6 D: u0 Q
+ B  U/ w6 @/ y: _7 f+ Q
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:* w7 f/ C8 q& G2 C
it is too easy:7 a, n* M0 H" b: G8 M/ B# |

! C/ O* @+ o2 v$ a0 V& Qd(a+bx)/dx = b, F4 {# Y" }% c7 k% _

7 _& R( T& N! z+ h3 Yso3 H/ F* O# W4 E4 G, q$ f

?' C: v5 ~( Zb=-kc+s1 |# Z7 p7 B/ J2 x4 @3 @9 C2 ?
- O1 H1 i3 y9 F. V% d( v
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:
! I) g* S# L: R请教大家一道微分题，多谢！4 H8 Z! s8 m4 Z

  E( {9 B# c: J7 xd [(a+bx) ] /dx = -kc +s ) ]5 N  h# p: m+ F
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
/ w2 }4 {) r2 M/ ~1 f0 Z" ^1 g1 ^8 n# m- p. U. \8 M  e1 _
多谢了！. N' l/ \! t6 m( f+ G# ~

0 R% |, c$ r- o7 W: a# u7 _/ F[ 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:
# X( u/ H3 A* ?+ R) P' Ssorry, did you post another question? your statement is still not clear.
2 A/ X* y7 @' p; }& b
& u( `6 T+ j" H, Pa, b, s, k are constants? c means c(x) ?
) j _# f3 f: [; V. K6 }3 u: ^/ i6 W& F
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:/ u1 K7 E. m+ p: a2 g
对不起，写错了，忘了变量c,应该是
( Z; A5 R3 n, }- n$ ]d [(a+bx) c] /dx = -kc +s 7 F$ t% c% I" Z1 w9 M# N
9 {6 y/ Y4 C) x6 h" Q
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:
" _$ z( b& x% y& M5 g! ]6 \6 ddo you mean it is:& i0 D7 K0 f2 A$ p( {

w; Y. T3 r$ T; Hd { (a+bx) C(x)}/dx = -k C(x) + s
5 I* n. }* o) ~) s7 ` B
/ r3 Y! h& A8 \! Xwhere a, b, s are constants, you want to know what is C(x) ?! s/ s( n0 u1 \6 W

& j: w* Q) m0 _! f: aif 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:$ }1 K, M" P2 B9 [1 u O. s' ^
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:
; H8 h2 A% C0 }6 p4 D, I" m9楼的答案有点问题吧。
% a  r! D/ a6 R- J7 A$ d& `. Z您说：
/ {8 |* M2 j3 k( p& y4 r* Y) h/ O; Y: y1 a4 }" {9 `% }8 @
d{(a+bx)*C(x)}/dx =-k C(x) + s
& p& s) m  K! q! f1 B4 s5 vso:7 j. D* ^0 m- i6 V
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s8 ^* `/ G& d5 s* ~" W1 V9 e* u

, A( a& x7 ^0 B3 D1 ?. M" K" E& J4 t也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
" @6 ~) Y! I! p8 {  L% |, v5 e但这是不正确的
/ |  ]0 y( L+ q1 ]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
) b6 I& M- ]) r0 I+ I....
' O% d) @- r! V. M1 a, A- g, `1 Td{(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:* H6 c* p1 j8 M9 y" F+ a
这样算混淆了微分和导数的概念。
8 K1 y, B% U) E" C2 {0 {+ Q8 I/ R* n# `. w9 C' t& 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. s2 V0 \ F% \# C
& D( w9 \5 A# u: e
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:# X8 ] W9 u$ F0 L! C
这样算混淆了微分和导数的概念。. I# n, c# O& `( A1 c2 ?

d, }" C3 H6 Y" q4 hd(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$ z6 p4 l9 Y: q. N' c
$ t5 w0 R4 e% X
Ah, you made a mistake: it should be:
* N3 q4 F! Z0 \7 ?& Q# j; @/ W( y9 G6 S; @, U9 N. ^
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:
0 ]0 E" `! Y) V7 ]( I# {9 p" O5 qYes, there should not be " ' " after "f" and "g", it is a typo. :D6 L: ^: G, r7 ~! {. k
% R& @: ~3 i8 q3 M7 \; M+ O
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:
% b5 X) H; ~8 B# Y
! _. X! a! i" B/ i( w6 X* k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。
+ J9 Q* d* J$ ^( m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:
8 j* E4 a. M0 n& t; H2 G6 ~2 \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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