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:
) B6 q9 s, d5 l0 b请教大家一道微分题，多谢！8 r8 k2 x/ }( r7 C( ]9 @% R+ J( v

, V" ]. T: u, ?6 O1 T& e$ H9 L3 Q$ Jd [(a+bx) ] /dx = -kc +s
7 D' n" w( \5 g. n! e+ h. `where: only x and c are unknown, others are all known, requiire c = ?
, A" G! c5 o2 N) o) H# G7 R& r$ l3 A& b* G4 d( G3 Y
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:3 ~: C" L1 ^3 H; s: |5 e6 p
it is too easy:% Q1 \; i0 g8 d1 f& ^
1 c0 A% k- G/ V
d(a+bx)/dx = b
/ ?1 J) s3 t! h+ A: M) u, P, ~7 H/ I. V2 K6 v, A
so
L P7 z7 O& M/ y
) f! y3 ~# l8 |b=-kc+s
4 T# n& g, ~: o
& l, c2 L9 P0 g# X' gfinally: 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:
) h# K% v9 j3 x1 f- Z请教大家一道微分题，多谢！7 o  z4 o& t; o/ Q" b

- b% O: z+ I% _% S! Qd [(a+bx) ] /dx = -kc +s + L# L! D' {8 Z6 _
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?6 Z/ ~5 l5 {! n7 X+ }3 v/ Z

: g+ ~0 R7 Q  V1 s多谢了！3 _) }7 z: a, ]# O8 M7 R
9 L9 ^5 j  O( H7 c! R) q/ a) T  V
[ 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:
( a, V- T1 r3 Y- Q* v, k& q7 ]) Lsorry, did you post another question? your statement is still not clear.
+ j8 o' I( S! @! G
! h5 D) F5 B2 c3 g8 }' @% wa, b, s, k are constants? c means c(x) ?
, _) l0 s' T0 I
% I) F) T" s5 \9 L+ j; }7 uplease 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:( O& Y* K2 {: L( E7 ?9 \, s$ Z/ G
对不起，写错了，忘了变量c,应该是
9 O2 B1 E$ D# Z R+ `d [(a+bx) c] /dx = -kc +s " c) |7 Y/ {- U5 T$ ^
0 X: j a" \. c7 `: T) i
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:
! `0 N( S% j4 R& w% X [5 ~do you mean it is:
: {. B2 L% D6 Z( V& m0 U7 u
" o3 K2 |) g% t1 ^d { (a+bx) C(x)}/dx = -k C(x) + s
6 ]/ [$ s" ~8 ~. ?4 a/ U6 g
( y) N7 G5 i: b; j$ Y$ Gwhere a, b, s are constants, you want to know what is C(x) ?6 |7 H+ M# q$ K

! G0 H9 I3 s1 t; Y rif 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:
0 x w" c" S5 W) X1 c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:% W. O; q$ G; V" ?
9楼的答案有点问题吧。& d6 U4 q6 C. R7 E0 ?% x
您说：7 H% @5 ~, h$ U8 K8 `8 }

4 {1 [ V8 C1 o Od{(a+bx)*C(x)}/dx =-k C(x) + s9 B; o7 \* q# t0 H8 [# b
so:
2 j' f9 {; @. p5 Z$ l1 U5 w" A9 ObC(x) + (a+bx) dC(x)/dx = -kC(x) +s' A( j7 k- \" j) J# G" ~' _

4 w3 n: m' m0 ~9 k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9 W$ i& A% Y7 C$ H/ t3 v/ `
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
6 O( p2 }; g7 s+ w....
/ V9 u' M! g0 v( ^6 Cd{(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:- {/ O& u9 Q# v, j
这样算混淆了微分和导数的概念。, D. `+ k! W9 ]0 d! ?% \
/ p, y8 K7 {8 h' y- x/ r* R! }
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
# l2 b5 Z/ n e
1 P; T4 S0 V% @8 ~2 h# K4 m! i% f0 aIf 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:
2 h' ?( G( u# K1 S7 Y这样算混淆了微分和导数的概念。
+ r$ w; h9 L8 `# U9 y+ s9 l! B& Y
" y0 {6 [, j( R- h% C# C3 B8 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3 ?( u! P5 \5 S" I/ q) L
8 c5 D/ @9 F! C! {
Ah, you made a mistake: it should be:
0 f5 F6 O- Y: q0 v( F) B1 Q2 w
# k0 Z, p5 p; y. r* t. Kd(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 20:51:2 Z/ M( F+ k6 z
Yes, there should not be " ' " after "f" and "g", it is a typo. :D& U9 y6 j& g. w" ~- g1 t6 ^+ v0 P

5 L2 g! F, u/ x1 S. _. [; OBut 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: [2 w2 ^2 G2 G6 j8 F! V
- j1 Y! g: l: G) ~
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 |# @9 u! H3 _
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:+ q0 ?5 |8 K& o, W- I# |
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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