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 20:41:0 m( J/ `' l4 C) B h6 _; P! Z7 F
请教大家一道微分题，多谢！$ r1 v: a0 @5 E1 @4 R
% @8 O6 H8 G% E
d [(a+bx) ] /dx = -kc +s
; l3 S4 j. A) a# Y( ~5 Swhere: only x and c are unknown, others are all known, requiire c = ?5 E) P- g- [! y" R- O$ y

' y+ W, `( O, J7 r) f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:
1 q$ S' d4 U8 y1 O( y) @3 kit is too easy:
# h- _9 c$ a- [; \0 ]6 z# C9 g( I2 Z" A
d(a+bx)/dx = b
. N) \, N; p% C2 }3 K2 V  G( w0 S0 T* B' i- @, U
so
$ y; W. }# h! E2 x4 ~" o) }  r/ T8 g/ \% @
b=-kc+s
: L1 C7 H4 m: {! k2 D6 B- V6 y* ^" l: A
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:' P3 s: Z' ]( }' N3 {
请教大家一道微分题，多谢！/ `9 O% l1 S3 b) S% s
7 r( H" W; @7 U
d [(a+bx) ] /dx = -kc +s 3 ~9 I) D; }0 I* O
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?! l3 l$ l' R- _ p
' B% C6 R) U6 r8 f1 j9 O2 f
多谢了！
& D5 `7 ?: d/ R% E0 g7 ~! I9 i& W( h& }- d: W) q
[ 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:
5 O3 i' T. x4 G& asorry, did you post another question? your statement is still not clear.
9 N, B/ y0 @$ \$ j2 {& b( C# ]- f9 p! e$ X, Q- w
a, b, s, k are constants? c means c(x) ?
$ q! R+ x) n: f5 J( p& e& V0 y3 m4 t& t# Z
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 n" ^) I2 h1 c) y% ~5 U对不起，写错了，忘了变量c,应该是
" N f1 z- o: Bd [(a+bx) c] /dx = -kc +s 8 A9 B8 S+ y9 s

9 o" i3 z, z' t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:- b: z% D# ~4 g0 G8 x) S: ?9 g' ]7 A
do you mean it is:" R f4 T% ?, _' U2 d1 A
% h1 ?& W4 q; U+ S5 Y, H6 E
d { (a+bx) C(x)}/dx = -k C(x) + s
/ s! |1 y% \7 g2 J$ X
( k! k- ^! w! R6 X5 V3 K1 S' S) f# awhere a, b, s are constants, you want to know what is C(x) ?
6 C6 W F, J; U" A; x! V# _5 F- ^
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:
+ t, C4 W% @# a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- N3 g) G( `
9楼的答案有点问题吧。
Z/ J0 b* Y, b. A+ v7 \( W4 W您说：% i) u7 s9 W! d3 h0 K) s; J

% P3 u& j+ v4 P: ]+ ?) Ed{(a+bx)*C(x)}/dx =-k C(x) + s7 x7 o$ ?9 k) _$ q! E5 Z. u
so:! r9 b7 U3 f" k
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s/ _: g+ q# P5 c$ q0 w, _# l: b
6 x# M) V; e/ R; z7 o
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
# d4 g2 W7 ]) k6 Y/ H% k5 i: f但这是不正确的. T3 o2 y7 v3 H+ g( ~* I
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8 T. d" o3 E& d
....
2 c/ l L( p9 Y4 [3 e4 _0 e/ Rd{(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:
% P6 h( w9 L; B; c) u# i这样算混淆了微分和导数的概念。
; z% Y f3 V% G0 A1 d6 \% p* ~3 }$ }
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: k0 u3 X) Q) L3 o3 t
7 X, Q0 S8 e- U. D& [7 \0 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:$ O* b1 h6 U6 N+ _4 O- i, k
这样算混淆了微分和导数的概念。+ }1 I2 f" l! P2 _) p

4 h) N% n9 B0 f. b' c7 gd(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
* b u4 b1 a! D2 ^0 Y& d6 w7 j$ b3 S7 k$ g0 y
Ah, you made a mistake: it should be:
/ f3 G( I: a0 ^6 F0 b
; H: p3 g$ G; o! ^; Sd(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 20:51:8 z" h1 Z5 V/ Y5 h$ v1 d& M
Yes, there should not be " ' " after "f" and "g", it is a typo. :D. g8 K( V% D* M! T s6 ?* w7 x
- r7 |; H/ M0 ?5 }. Y1 t
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:$ `$ T8 A, V' n J0 B" h

) f1 D& S9 n3 R; w& Q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 g3 c/ S5 ]6 E& x
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 23:24:2 M f4 |7 h: h k* }
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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