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 20:41:1 L& g8 C! G9 x7 p
请教大家一道微分题，多谢！
3 m* x E& }% O; }
3 t# e! U8 ~! k( ~1 R- X( Qd [(a+bx) ] /dx = -kc +s
3 L, k/ f$ [; ^1 t7 zwhere: only x and c are unknown, others are all known, requiire c = ?6 n9 q% Q/ \% f5 }# z

( V3 R* X& w7 H! s1 _+ N! c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:+ a. k6 D1 `* N0 @2 H3 e U) J x4 K
it is too easy:4 A$ C) x( v) v2 ?7 W% O

1 |& P' v- s+ z8 {d(a+bx)/dx = b
+ o) a" [+ v& N7 m$ M
1 Z7 _/ ?& F" }so% n' U9 Y( f. [2 d

: Q; R: h1 E* {3 fb=-kc+s7 q- T+ y7 i" K# ^! f
5 H% ~: |' B8 O
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:
- f& {$ e# H" F" s, z( r  K4 B请教大家一道微分题，多谢！
9 U3 i: _! q3 ^8 W4 |; s( B+ j; u$ u- V; M+ B# t0 h7 i
d [(a+bx) ] /dx = -kc +s   e. J( |; x6 `
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?6 z  \; k5 I- `' r9 Q' p9 P
, h) J  ]  T3 J" f3 Q2 G
多谢了！' x$ z* o& e8 I2 q1 A

1 q6 M( Q9 m; P; l[ 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:
8 G. d: _7 x) y8 e1 Y- z5 i# Msorry, did you post another question? your statement is still not clear.
+ |7 T5 P$ A9 P4 q/ x, \" Q+ a8 ^% M2 S2 H/ w6 x/ N
a, b, s, k are constants? c means c(x) ?
! g z- I, D! z3 D9 q, w" p$ @# p+ n# i% u/ ~8 a, X2 y
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:/ l9 E7 n5 E" o9 R T- P
对不起，写错了，忘了变量c,应该是
& A0 r4 S* g( J# }2 R, }( o; zd [(a+bx) c] /dx = -kc +s
$ i& A( K- y8 F- v+ l
8 p) n' z/ I( @! Z) q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:
t( C g1 v' c$ e3 Sdo you mean it is:
) d ]9 |9 L" |) C$ J4 q9 Y( x3 m: X( L# Z% J
d { (a+bx) C(x)}/dx = -k C(x) + s
3 ?- F1 w+ W6 a' d8 n- P5 R$ o) V0 O1 R( W( z& X- K
where a, b, s are constants, you want to know what is C(x) ?
0 R9 N" y6 H: }+ s3 G0 ?( R
4 J4 `3 F2 O7 Tif 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:: Z% ], t/ S& B
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:/ N8 T: b2 X/ T( ^! D5 P
9楼的答案有点问题吧。0 v2 _9 J7 i! p
您说：/ H1 c1 H9 z$ I1 \

* h$ X% H" o ]2 @5 O3 n: j1 c* z3 Gd{(a+bx)*C(x)}/dx =-k C(x) + s9 L N, u/ b! C6 l1 f% p3 p j, A
so:! N9 X( J# G+ j# m. q! g5 [2 a7 O
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
% v8 |8 Y$ d0 E( ~8 k; i5 ?3 s+ ], a% h3 `
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx4 V2 W4 q3 u# `4 m. [
但这是不正确的) o" i0 X+ a5 e1 e* S
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+ P, x( ~$ s0 i# V# y9 O) }
....
3 h5 L3 L" C* K5 ^$ z6 Id{(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:- t. Z# Q8 E' b: I0 T
这样算混淆了微分和导数的概念。
" ~" d: {+ X: E; |
: k& N J5 |/ Ud(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# `# Q. l( Y# t8 k8 V9 H- ^, Q
) d8 T+ y/ {4 b6 U8 Y# l+ t+ m' f
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:
( d' ^3 m# ^( x, t. S1 y% d这样算混淆了微分和导数的概念。
  N+ M% _: H7 O+ l) v  a9 v& W5 Y
6 f) z" D* l; x8 j/ ?6 t7 Dd(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5 ?$ r  _3 x, g3 T+ p- c9 ?

+ O6 p  x! u* n( HAh, you made a mistake: it should be:2 X0 P/ W8 ^9 s4 d, o$ j* R
- C4 U) R% O+ K% [. d& ?; O4 n
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:( E- n2 H U; U0 W3 \6 V# b
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
4 E: Q# ^: ]% P2 r) Z+ ~& |5 L* G, h% X1 }4 I
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:
' I+ H# b& h/ o! l/ M5 f
" n; F+ x6 v% W1 G/ V% I4 Z, a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。
) E3 c# h6 p9 j+ n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:
4 J0 g: a. L! N+ L$ Z' O) p# y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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