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 20:41:3 b0 B3 |& n1 L( r' \
请教大家一道微分题，多谢！
: S+ T  W  o7 L6 c; m. T8 v$ u& e
5 i  F* v8 K: |% f3 Kd [(a+bx) ] /dx = -kc +s
, a2 \( Y+ @& h" O" dwhere: only x and c are unknown, others are all known,  requiire c = ?4 R6 X. V  z, }$ D  C) {3 t
/ e* _' i1 Y7 H8 ~) N' r8 c' R; q! R
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:+ m, }1 Y! |* u2 w3 ~$ c' z9 d
it is too easy:, a0 U* e# y& G9 K

+ w/ A" p# V: H9 H5 xd(a+bx)/dx = b1 }" z4 i$ u* ~ g) Y- B
' O, ~) G/ {* ]# d) `- W1 `
so
" W: R# R$ x8 H
6 u' {: D/ i9 J( {. u* ~b=-kc+s
( B4 p9 i' h1 ]* q1 X2 d
) @7 {# o: l* Vfinally: 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' Q. J) P请教大家一道微分题，多谢！
5 o, w. c+ j2 C9 x% `
& M* f8 b% S7 o- ~+ h9 N3 r1 xd [(a+bx) ] /dx = -kc +s
; l' i, \/ J, G8 `+ j5 D3 a2 awhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
Y' g3 `. y. ?& T1 O. Z( A$ x. Y+ G9 W( v0 O0 M
多谢了！
8 N: R- K# e' l" |7 x* r, m, P+ ^7 Y6 R) J% i0 R2 Q. T1 I
[ 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:+ K: ~( V' S; V( f* Q+ y6 Y
sorry, did you post another question? your statement is still not clear., c  p. ^8 _9 ?

. N) a) T2 C" P; J# L5 ba, b, s, k are  constants? c means c(x) ?
* E' d1 n4 R; w( c6 ~  D5 l* o) j% f  ^5 ?2 t5 y$ k$ T6 b
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 S5 f; V; S( `! A对不起，写错了，忘了变量c,应该是" ^# p5 n- `9 I8 Q) z
d [(a+bx) c] /dx = -kc +s 6 H* D, p1 n3 K6 e' d3 p
) {/ R/ c. ]. M) t# e/ j& U, `
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" s) ?3 {2 c6 H) ~* Mdo you mean it is:+ ?, r1 k7 A( B/ i) ?, T
7 A. P2 \9 W: P: Z$ `+ Q
d { (a+bx) C(x)}/dx = -k C(x) + s1 c3 \- f% @: S" k2 M& t/ o2 D
4 `8 e* F: D" j% q% g
where a, b, s are constants, you want to know what is C(x) ?4 o- `8 f5 `4 {( I6 Z
E* n% z4 I/ L/ ` j
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:$ X- h9 P' g# y7 _- ^' p' V( u4 } U& ?
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 19:34:6 I/ S1 c+ Z3 `; r
9楼的答案有点问题吧。
7 z& U$ J& }$ [" A6 _" V. X您说：
3 m, C8 y: A* W- o% `; B& w" t# A
d{(a+bx)*C(x)}/dx =-k C(x) + s0 t' M: ]. k5 Z& A
so:, E/ Q( |4 p2 R; ?1 m
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s( \' U, V& E* V' Q- H

) x: Q& P. Y+ O2 v0 K8 C- U5 B2 S也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
, g3 r2 _! S' f; o& W1 l w( [' |8 K但这是不正确的
# r! x+ K4 c7 w1 \0 Od{(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
$ x: S( o$ X% {....# q+ U+ C0 K7 N0 o1 R1 {- f
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:
$ v( X- p- g1 m* z. e2 e这样算混淆了微分和导数的概念。
+ D7 _0 ]# h( d9 G1 \
; K6 R8 e- c7 f3 c4 x. B5 Cd(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
9 |: R5 n1 _' A' a G& c! }9 U8 C" k$ t9 k1 q
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:
+ q: D8 y e; D5 y$ y# |8 f这样算混淆了微分和导数的概念。# z2 x% T6 Z% l6 v' \

& t8 X( ^* b* a$ i% 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
, \, P. `- Q4 U# ?* K6 a4 M* T! }* B' K, l) ]) a" f
Ah, you made a mistake: it should be:
5 @) _/ _" |) t
3 \( W6 [- {; S. J0 P$ ~0 Y: D& |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:
' d) s4 B) ]3 e$ k4 y8 f* lYes, there should not be " ' " after "f" and "g", it is a typo. :D7 K- V) E! l& w5 c

9 b+ C/ ^/ {3 Q. Y3 WBut 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:* R# H" X% ~$ P @0 f' I3 X# _

% k& x2 z1 U' D* \- g7 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。
' m+ B3 x) A& T0 }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:
5 m% l" x0 b6 T. f7 g0 h' x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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