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:
9 g8 L' y0 a9 E- W8 }8 `1 \3 L请教大家一道微分题，多谢！* G, u. d( I. U4 w1 X' Z

5 |0 v4 ?5 K7 gd [(a+bx) ] /dx = -kc +s
2 m2 P1 r- _) {where: only x and c are unknown, others are all known, requiire c = ?
3 {1 K8 O. r5 I; Y5 Q, Q. H- Y
; ^5 Q- @# M1 S# ]. W$ R/ 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:4 ]6 e  j5 P+ @
it is too easy:$ K( t- M7 j% }" w
7 x: g! [1 j4 W* o  [: V4 K- o/ m
d(a+bx)/dx = b- p: _; p( w* r) L
9 }9 |9 h9 h% G
so
( C9 S( S% F$ r$ u
, f6 e8 E9 F  a" q7 w* U$ H) Z! ^b=-kc+s
4 a$ y6 ~0 y( M. P) M& e5 w) M7 ^4 I! {7 E
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 20:41:5 \5 t {2 r: R
请教大家一道微分题，多谢！" \; u6 Y" J: O( v8 D# g5 _6 ~& L7 Q

2 r& E$ Y. U2 F! e/ Hd [(a+bx) ] /dx = -kc +s
+ |# R* [# O& L) o! N. pwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?* n, w5 m6 d2 i+ d* L9 l0 T$ I

- b* Y" N; \, o8 {多谢了！
) L, m7 L! `9 R& V
/ `2 R* q' h% P[ 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:
  |: E; Z6 p; ]9 ^sorry, did you post another question? your statement is still not clear.
( t6 K' t+ Y% z1 ?6 K& h4 v! r( \% \
# r, F" [$ Z4 a+ {& R0 ja, b, s, k are  constants? c means c(x) ?2 Y! d4 @* o( _

, @5 V; F0 D7 [. C" Q2 h1 y  rplease 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 12:30:3 l5 U+ P8 p4 g8 ~# w
对不起，写错了，忘了变量c,应该是: s) Y+ u. Q9 Q/ t- d3 }( }" [& Y
d [(a+bx) c] /dx = -kc +s
' J" K7 r; M" B5 n
) c: _1 m' g2 C, e3 A/ b0 \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:
* s, d; \6 G/ h( B9 a) mdo you mean it is:' \, w' W$ T* u% q

5 k$ g# w/ ?( F; W+ i2 Sd { (a+bx) C(x)}/dx = -k C(x) + s
* e/ S7 ^+ S2 U1 U6 N, r j! ^( ^$ D
where a, b, s are constants, you want to know what is C(x) ?
0 f1 K c' \0 i3 Q; q* a2 Y. Y) }# }1 `! R2 ~! _5 ]
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:, p! h' n0 [5 O8 i
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:
4 c! n) [& l$ L& _9楼的答案有点问题吧。5 d  X, B& p! ?+ V# _
您说：
- J0 X+ Y2 K: L5 _$ L" t) A* [7 r, s6 t  g- d4 d
d{(a+bx)*C(x)}/dx =-k C(x) + s
5 P! w! S6 [1 S, gso:' r& h6 D( c; T8 b! o. {6 k
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
! `) z" Q: a% o
* @3 O: U: B4 Z1 x; s. K$ b! Y( s也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
4 {$ G- [& x8 d- c. q6 z" l但这是不正确的
7 o6 l0 T# z$ W& P1 v! 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" x& A w4 p+ S2 G3 g2 ], l" L
....
3 m1 U) |: }* z: wd{(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:) g2 m7 e1 o5 {
这样算混淆了微分和导数的概念。- g2 n4 b6 ^: l5 m4 C

$ |9 w8 Q2 [! P9 b: x) Id(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
$ Z9 O7 N. c5 }) x" y ?; B
5 G) N7 {: `3 R$ J( a; K3 rIf 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:/ a0 U2 O& `; x% b; e2 ]
这样算混淆了微分和导数的概念。
* T2 k  K. b/ S' }9 V6 C
  P. o/ B" L2 ^- I! Z1 u) p9 P. f( vd(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
  R: |3 Z9 E* m
5 k. G  \. [/ @5 }. O+ u; H7 kAh, you made a mistake: it should be:! S7 J0 b* E5 t; t

) T7 @9 f. N* k- I5 B6 F; }9 Ld(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* W! @+ ]" J* F! a. g; j7 r
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
8 h# Y3 Q9 N% @( N4 p4 D6 l* S4 v8 P. K. g8 {
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:
' _: \* f; x: r
! n2 D/ t4 Y2 k+ r& p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。) [. y% i1 w3 I; b0 t @
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:
d- ^+ M- j) 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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