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:
8 B. x: J4 J( I+ J+ R7 P请教大家一道微分题，多谢！, F" D* @* |* q1 }7 [! J

. O( u4 K  [- U0 j3 |  d" ]d [(a+bx) ] /dx = -kc +s ; q# w% I& k$ Z) a- @( y
where: only x and c are unknown, others are all known,  requiire c = ?- n" f8 \5 n8 t' v* `$ w

* F: N' z( ~$ r' ^& x  z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:% d8 q( J/ {" d/ m2 }, K
it is too easy:
' N1 F4 C" t$ l
( Q* D# Z- b8 E" Xd(a+bx)/dx = b" n+ i4 H& k* m0 t1 q5 |" z( t
- u7 v2 ^, c' z' B, S
so" n1 j2 l" {( @4 I( D( s
2 T4 N, ^3 G7 `" A) D2 K3 |- ]
b=-kc+s
9 V0 E# w' }- t' D) t& y) M6 Z; x& s4 k+ w+ Y2 u; f1 I
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:" s$ F* c; k7 s1 U/ Z( D" b
请教大家一道微分题，多谢！
$ C% m; x" a6 o' |6 C* h6 e2 ~! r+ {
d [(a+bx) ] /dx = -kc +s * S! _* `: t4 ?5 r! o% }
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?. N3 _4 i* {( V; U8 A
# y- t% M& F6 p. p
多谢了！6 `' @1 u' ]% [: h

* G# X9 k' M6 r[ 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:
" o$ d' l. r) `; Osorry, did you post another question? your statement is still not clear.
1 w, ~. M6 w/ B; x( X7 s$ h3 M: e) a  k. R8 J' c% N
a, b, s, k are  constants? c means c(x) ?0 D7 o6 Y# T  `6 D% x

' C7 r: L; |; L& g  T% 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:' ~$ f( P7 `5 y& h) S
对不起，写错了，忘了变量c,应该是) Y( ]8 y4 C% [5 Z. M( F& E. p
d [(a+bx) c] /dx = -kc +s
* H- m2 h6 ?' G& Q/ k% y( B$ D- { J$ N6 o! a2 F; 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 O C' _' R8 R; k8 |* Z& l; t6 M
do you mean it is:
6 s9 G/ e4 b8 J/ d! y4 R/ _3 S: }' l }8 p$ \( c% G7 Q9 [
d { (a+bx) C(x)}/dx = -k C(x) + s
% ^3 x; |: c: [% G
) @* D4 h2 j3 {6 X/ C. Z) c$ Vwhere a, b, s are constants, you want to know what is C(x) ?
' ~8 g5 q# O, E( s
% ~+ p) A& `+ |/ U# R6 Z" S1 Qif 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 d4 _2 }# H# ~$ T9 D4 C2 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:
1 p  {! j( N- R3 w0 k9楼的答案有点问题吧。
+ z1 y% `, p) ]! K* [6 d您说：
3 M4 Z1 u* [* b6 s! ]! M* y% V) o8 @: v: J. j8 v5 l  L
d{(a+bx)*C(x)}/dx =-k C(x) + s9 Q2 ^6 K! N: P9 d6 W; ~" Q6 A9 K
so:2 E: A5 _& u% C
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
8 c; p5 b7 ?: f
5 \: m4 w$ W! b5 Z也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
: w! I2 F# U# v+ y3 ^, R: J但这是不正确的3 [3 p6 f# ?; L7 [/ @. h
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7 v% ~6 E: \; {( W! a2 o8 c9 a
....7 m q+ d; O- u( f- Z
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$ O/ b2 {8 `* g8 z. @
这样算混淆了微分和导数的概念。' w% P8 O0 s. C0 `9 j
5 f% o! J8 l2 l s* 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
- g2 @6 k( E3 W ]) Z5 F3 X6 Z" c5 [ y: A/ c2 e, v* }4 Z) V; P: x
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:
9 e9 ~+ M7 n1 m1 }这样算混淆了微分和导数的概念。
+ x! s& f, N9 |3 P( h) X0 Y* y! Q" @6 _" F
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
! b9 n4 d, D. G9 Y
* o% }4 G( y/ e' o8 \# QAh, you made a mistake: it should be:5 q6 y# b* {! W' Y9 `, N4 }4 n
8 f; V, g/ a2 n% Y( ^! 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:+ }4 m, v' C2 E* n+ ^( ^
Yes, there should not be " ' " after "f" and "g", it is a typo. :D9 c% \4 j1 g4 c U% w. L
! i2 ?1 R3 U1 D6 Q- E
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:
, w& x5 I1 E2 U! q
- Z+ U( Z; n9 W8 b5 k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。
, P: e5 F6 H6 j9 T8 h( h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:) P/ N2 k# G. _* X( P" U4 P9 l
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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