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 Y- ~4 y! R8 o' [请教大家一道微分题，多谢！3 _7 u9 x+ \7 [( \! m0 c: d
! E5 G: P( k& t+ w8 L/ f3 t. a
d [(a+bx) ] /dx = -kc +s
- K" S1 i& u- H; `where: only x and c are unknown, others are all known, requiire c = ?
+ T' v4 P% D; W( ~+ c( R& \0 c! ] H4 H1 T& v. M
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:" O5 v) Y; ]. Z, X
it is too easy:, \6 @4 E6 ^' K9 m5 V  l  m
$ Q" z! u( `$ k! Z/ B
d(a+bx)/dx = b
2 ~6 ?9 ]. Z( Z8 K; E+ g5 z/ p( p% i5 k+ A
so. O  H1 p2 R& }7 y# r$ L
& |% _. V" L4 |
b=-kc+s5 P: o9 s3 O& }! J& D
/ M" |. u- e' ~+ l; E. F+ n$ z3 M, E" m
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:
4 J- k/ r( |2 {* b9 m& y6 r请教大家一道微分题，多谢！5 `2 L- S0 Q. |1 e5 C; }
. z* N3 I! L) }! X* e, ]6 d
d [(a+bx) ] /dx = -kc +s
' T$ f3 I! U3 z) Fwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?) i( j& i+ ?/ N' {
+ w1 d& x! A. v" a) h! J
多谢了！0 c2 C6 Q, Z; ]) a6 Z+ t& i" S/ e
3 B2 T e( a3 E: u$ s$ I$ [2 f
[ 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:4 J! g  L2 o% X9 M
sorry, did you post another question? your statement is still not clear.
* Y) q& d1 w9 o( J+ _. ?) c& S' F6 m* l  i/ z2 l2 T: K
a, b, s, k are  constants? c means c(x) ?
4 ^! A, d* D# d# m$ B* ]
4 K2 A" I/ J* i; M5 Mplease 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 12:30:4 Z3 f' }0 X' D4 g) `- E/ L5 a
对不起，写错了，忘了变量c,应该是
0 a) f, ]1 f, nd [(a+bx) c] /dx = -kc +s & j( `& ~, W5 v9 d3 {3 B
& X: Q( {$ c2 m7 v, v4 I1 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:
4 s9 f1 A% P" s2 ~/ g- Sdo you mean it is:
5 P8 i- _* @6 ]+ Z$ A2 M. `) y% T' J% r- p3 u
d { (a+bx) C(x)}/dx = -k C(x) + s
& `) n4 \3 \: t: q# h8 u, Y- Z8 c3 O9 ~' K a8 U' M! Z
where a, b, s are constants, you want to know what is C(x) ?
4 w6 r4 Y6 u& d2 {' ?
" S2 s; V3 w6 E! Z$ e: L9 b% C5 @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:) Q2 D9 H2 G4 O X5 v: ~
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:
+ _! i2 o: N: v: L+ m9楼的答案有点问题吧。/ W+ [, p6 }% i
您说：
/ U; f# A6 J: ~  K' E+ Q+ z1 d- g
0 L8 I* Z4 x* _! b7 o) g1 ~9 qd{(a+bx)*C(x)}/dx =-k C(x) + s
' T8 @" U7 k$ Iso:
9 n5 s6 a2 k" {* e* LbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
. a$ B4 n  }: G. I6 L4 V
; G8 {, Q4 D6 p* p2 _2 b3 O4 X也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
" \5 r( p, G: B& h+ I9 `: W) l但这是不正确的
7 h/ c- e7 O& v& nd{(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
3 a' N/ @3 J5 Y2 h: }9 j1 @....
9 Y( j6 k- G- z- s! w0 V1 yd{(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:% W( ~6 a! h/ i* r0 j% n( Y4 p
这样算混淆了微分和导数的概念。% ?$ [ z% J+ f+ |0 B0 |) w% N
0 o+ u# P0 m+ s2 S$ h6 ?
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) i% I/ Y2 b& O0 W& ]

7 J3 D7 ?: q+ o( b7 n. J4 `" SIf 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:4 s8 {* L4 E! O3 H" ]% C
这样算混淆了微分和导数的概念。3 F n* |. x' G4 D1 r1 A
$ N" Z) U. h+ O
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/ I! c; x. n( O5 k# {
7 r% r* Z e' S+ L/ G% z8 ?2 J2 S# H
Ah, you made a mistake: it should be:
4 d" O7 O( _3 A. L% b2 S" C/ P% x; K+ S9 z1 W
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:
# w3 C" T5 Z' a8 u1 R) t: GYes, there should not be " ' " after "f" and "g", it is a typo. :D
. c! g a1 j$ n
7 z7 F) k6 U( h- ?, NBut 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:
: N6 P$ Q8 T" [: f% H" E J) b' p$ y0 L6 a# w0 Y
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。3 z0 n) s, a% K
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 23:24:1 A, J) X. F. C: S: P- E
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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