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:
) P# x' B# [: J5 v& E5 [* b6 v请教大家一道微分题，多谢！4 p3 i+ r5 c, E5 {4 {( X
+ N9 ?! `; y, X7 G
d [(a+bx) ] /dx = -kc +s ) |/ c& Z0 a* `5 r& E
where: only x and c are unknown, others are all known, requiire c = ?' Y: T7 _, L, d& p2 g b

4 o5 W$ T- L1 \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:
+ x- M+ W6 h' a- D, e: C0 Hit is too easy:& p' B+ `  I" {1 d- X" j6 M+ o  `

3 [% b: D, [3 \3 ^! {8 r# ]5 J$ jd(a+bx)/dx = b8 X+ K  Z, Z$ l' l/ V& c) Z, _8 }
& U/ p$ z5 U, s/ M- M( J
so
+ v9 R1 p% f9 g' y
- z' g9 h' P7 Z, H: Fb=-kc+s& G9 w* F# ^4 G! `8 e5 g
: ^( z9 m0 v4 x8 ~3 B
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:
6 A- x, i" w" v' C; C3 L0 m: Y请教大家一道微分题，多谢！
3 G) x% w: U# k4 ]  p, w; U8 z" ^# a# J' N% `- {# l6 W% c+ `
d [(a+bx) ] /dx = -kc +s 1 M2 w1 Q/ N- w) R+ ]2 |
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?1 q7 c$ L' U$ w3 b, X
$ f- p6 z( J  J1 s
多谢了！3 q0 z% p- a% _6 {# M, j
' Z% H# H; ~- N5 D: ]5 _
[ 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:& Z+ ~8 I  |6 h3 b9 o8 L6 [3 N
sorry, did you post another question? your statement is still not clear.
8 P' o- |( B2 {# X' ~% R
& u9 c5 v4 E2 b: @& E  d+ @) C4 la, b, s, k are  constants? c means c(x) ?, f0 G% ^0 g. C' b4 h# A5 q

, q8 n- o7 z5 N% m2 Q  iplease 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:
" Q, J' {" ]: k+ @' o对不起，写错了，忘了变量c,应该是
. ` Z: c+ B! id [(a+bx) c] /dx = -kc +s 4 O$ Z+ @! T/ Q& u* \" A

" d0 b- A; B% h& G8 m$ \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:
4 j3 j2 p, n) o7 K0 E5 k2 hdo you mean it is:4 G+ W5 H% [! [: |' y

! N9 x3 d( i5 \( r G# l1 ?6 Wd { (a+bx) C(x)}/dx = -k C(x) + s% A& i' `6 F% q" |3 d+ u
4 }1 S/ f5 W/ J) z- _; x& q" x
where a, b, s are constants, you want to know what is C(x) ?
, d- M+ d* c M4 u7 O: f. i' ^9 w' C2 o+ h4 m2 t( A, v* Z# A
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 14:33:8 w N' `9 i% ^/ D
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 19:34:3 f8 ^  p4 U4 W4 t
9楼的答案有点问题吧。
- ]# i# ^% x6 a9 ~2 J您说：3 r$ z1 A5 ]: I, B; j0 f
* p$ j& P% R* C" w. b' t) T4 h
d{(a+bx)*C(x)}/dx =-k C(x) + s
: |$ j9 }! m/ j$ p. x* x8 hso:! J/ V& S# j! v! [5 u
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s/ g- ~- [/ c: `# K. s

5 B0 [1 n5 J4 d6 Z; L. l/ ~也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
! V$ q2 P' W/ @6 L, l! t但这是不正确的
# w0 ]# }$ x% {% w1 C" u  ?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
0 \- Q/ x( }' Y, [: u....' }8 ?: e- M G3 k/ I4 B) \
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:( k& f$ `7 m3 O& B) V
这样算混淆了微分和导数的概念。
' ?' } n3 o$ I2 t; X2 B# q! x( e2 n8 m' A% g, d% T2 \5 {
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
( ?1 H% S: u, `9 z: Q, |& F. O/ T2 p6 ~
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:/ X5 w8 ]5 I/ I1 B$ M ?
这样算混淆了微分和导数的概念。
( ?& Y, g1 S) N/ a* q! Q/ B, c8 f9 c! F" f8 |+ h
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
" k+ b" G4 n3 _7 f
4 L. |, k8 e0 y% IAh, you made a mistake: it should be:
* {: R6 i/ L2 G2 K$ k0 J! [9 J# L" H; ^+ e. [, C" r b: E
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:
2 Q1 M9 M' i! k, U# lYes, there should not be " ' " after "f" and "g", it is a typo. :D
" R$ |+ |% ]; |( _7 O, V( `2 ~- a! I( V" }3 Y0 k
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:( p* B4 N. n6 l- F" c! b
% b0 Z2 A4 U( v9 U' l& m2 M
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。
9 i' J; q9 _% \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# j% d1 g3 |# E) |" `/ E4 \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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