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:
" t5 \2 ]* v4 ]- s请教大家一道微分题，多谢！
6 S6 C/ R0 J5 z6 {$ H$ w1 \$ ^; s+ n8 ~' N# `* F
d [(a+bx) ] /dx = -kc +s
$ r) |0 e# O6 j3 D: `' {3 o5 k8 Jwhere: only x and c are unknown, others are all known, requiire c = ?8 _1 g/ [& I2 N4 _; l- m+ }. I
+ P ^# }; C7 C3 X5 u
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:$ x2 T' _. y( [& y! d
it is too easy:- ^: R+ p4 R) R( M2 u' |
: t. @( }* }+ D! U8 u
d(a+bx)/dx = b9 u$ B) _3 d) `
# V' }2 Q4 C' B
so3 b% Q+ T2 R2 M4 w+ Y3 p3 [: V) l+ w
; Q- r8 `/ Y; b1 i/ k( y
b=-kc+s8 D$ l. n; a9 _) ^$ D& ]

2 A+ z. g- B. u1 y0 t8 P, u) jfinally: 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:( i/ u9 c5 G" s
请教大家一道微分题，多谢！1 b7 ~* G: E; l- v, D
% U7 e% i' G& _. m
d [(a+bx) ] /dx = -kc +s / ^! q* |$ E9 `8 P
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
' P% G/ R: X( f1 F
$ m% @2 q% T& Q& C) v9 W- ~多谢了！0 P$ x$ ]! v4 t2 d" R$ c" t! i

2 ^5 b5 q; h6 X/ m) @! i[ 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 M* M1 N) d0 a( u0 ]sorry, did you post another question? your statement is still not clear.+ k) ^- `* R$ e3 G) C
3 d) x( Q4 K# d7 N( T5 c. ~
a, b, s, k are constants? c means c(x) ?

8 [( K! }& T8 @' p! Vplease 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:/ z0 D+ _3 @ s
对不起，写错了，忘了变量c,应该是
f/ g3 v* Y3 h( t D* I& wd [(a+bx) c] /dx = -kc +s 2 A6 M' j. s# R1 g: r7 {

7 D' F4 r8 T# i7 E$ G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:
6 G7 n* b- }8 Vdo you mean it is:* G. b% P- R4 a0 g# i
5 i `0 V' p w U) x
d { (a+bx) C(x)}/dx = -k C(x) + s, b# E3 }/ T: i2 x0 \; u

+ n8 z3 g/ ^4 Z+ Xwhere a, b, s are constants, you want to know what is C(x) ?6 y: A! W1 L9 T+ U- e' X

3 N: v& J, d9 ~3 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:
2 u9 F; ?/ E3 I0 I0 ?: ~, q9 A# b0 x3 z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:
; t  i4 @  ?1 L9 y3 y. N" O. h9楼的答案有点问题吧。
& B) ^! n; N6 ~您说：
1 ^& k7 P4 c* H: Z# P
/ l/ d/ x6 @% G  E* s+ M/ i& J. _' vd{(a+bx)*C(x)}/dx =-k C(x) + s& M+ g5 C% Z  H5 [; J$ ]- p: l6 p
so:
' E1 L* f; O' V- ^3 |8 i' d5 I5 PbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
3 d9 f+ M' t# @
% D' z- b  v5 f2 ~& O# v# @也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
, X7 ~. ]! }* R4 ]# W但这是不正确的% f* h; a# m* m0 Y8 Z; k
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
) n2 m+ y; y( a....
$ }# k3 o( a; v5 u# n# V7 d/ V: Od{(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:
$ S* c& O R/ j: e4 a/ z这样算混淆了微分和导数的概念。( {( R9 q% i6 z; ~: H" k# F6 i

& H& S( i4 P7 ~8 G/ 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) U8 H$ O4 a6 r+ ]
+ A1 t! l$ V' S! \2 e% L1 z" T, 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* w0 U( ~0 L8 x6 ?9 N这样算混淆了微分和导数的概念。
6 F4 C. O- q2 t L( f( Z: v+ G r- o+ s5 p/ A- `: q2 S7 z
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
6 H4 U% K' P7 L w. J5 P: W' c+ V8 P8 b' |
Ah, you made a mistake: it should be:- A, x; V# |# X6 m

& v- M2 M7 U3 e' Zd(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 20:51:6 W$ Y" t& r! a1 c" s* @3 e
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
7 L! v1 x7 U; i3 S: T" @' Z! k$ p0 L! t
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:# m! w+ }9 ~ ]9 B
9 ^3 ]( B! |0 @: v) a/ 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。
' A* [/ q9 W. r& E( p8 g3 L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:) ~- t* r4 y( ~# ^9 @' R
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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