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 20:41:5 H0 h: q! m/ c8 g
请教大家一道微分题，多谢！
/ ~+ r2 Y5 L6 X, h. D' f$ G- v/ P! I: A5 w" R4 e
d [(a+bx) ] /dx = -kc +s
! ~& H# U$ u$ Q# h$ xwhere: only x and c are unknown, others are all known, requiire c = ?3 H" R$ o# H7 r; t
2 I5 U4 b4 T$ I4 K
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+ l8 d$ V0 F5 z7 Q
it is too easy:
7 ~5 k2 _4 U: Q: g% v0 N/ ~
& V4 V* N3 O  e( G/ ^d(a+bx)/dx = b( ~2 O9 R& r3 d* Y
( v4 Q0 g! b( V0 @* ~
so
; L, Q7 E6 H9 V4 G: M3 ?( w
0 `9 _* l$ v0 U& f0 Eb=-kc+s
1 e: i* |/ h" A0 X  R3 H, R3 \. M9 B7 R  Z, \
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:
: C4 {. U) p7 {6 C4 I请教大家一道微分题，多谢！
8 W6 s, @- v5 O: c" J& [/ G6 M! g @' E2 h& Y3 Y7 g7 r
d [(a+bx) ] /dx = -kc +s
! S& G; D: u8 _; _where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?2 }+ A3 V! _. H# o+ J4 f8 y
& }: L% d7 U# k7 `" n1 l
多谢了！# A! W0 G- X; O$ L

* V$ ^( P2 }8 B[ 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:
3 ^! j! v- E) Y9 D& y) w1 F' W, [sorry, did you post another question? your statement is still not clear.5 c; K" W) y$ P$ s' w" Q5 X" Z
7 e0 b4 z' y2 f+ q
a, b, s, k are constants? c means c(x) ?: G# m7 }1 p, P3 b, |% N- ]

7 f! Z# R5 A& p4 w, ]: yplease 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:
8 i' Z) s+ D' R1 N4 G$ T对不起，写错了，忘了变量c,应该是9 \1 @- G; ^# T6 [3 O
d [(a+bx) c] /dx = -kc +s
* D+ S: [ ]: s0 J. i) }( ?
3 r0 }# O* Z; j8 g2 H( r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:/ j* p( m* ^" m0 |: Z5 Q: _
do you mean it is:/ G3 U1 w2 m, k9 c* F

/ Z$ f! _) ~# ~d { (a+bx) C(x)}/dx = -k C(x) + s
1 U# u4 _( \% q8 N) ^* U0 k3 G
" A; F9 h3 S! w: w- l/ [( qwhere a, b, s are constants, you want to know what is C(x) ?
# R1 g+ A8 m' s7 v5 W
% D% B6 N2 _& D( g* X0 b) `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:
+ k2 @9 F- H- i; n5 c$ Y7 r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:* e% j  E' m8 Q7 L, V
9楼的答案有点问题吧。  [% `$ y& i8 Y- F4 d
您说：6 w  D+ h/ U3 u  r1 Q
6 k+ \- X% x, g
d{(a+bx)*C(x)}/dx =-k C(x) + s4 `5 P* m# U8 w/ @+ B
so:
: I. D3 Y. q6 {( zbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
8 h- D& b4 \7 F* M" Z( P: D/ [* O0 W1 T3 z) ^9 _% o; h
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
/ F$ I* w. S; X# W0 k/ D但这是不正确的
/ n" W. }7 [; ?( H8 _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 t4 p# j9 B' k: |! z' ^! I- G
....
! P: B" G# f" Y( B! x8 Ld{(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 20:30:8 f E) m+ h$ H2 n( N- f
这样算混淆了微分和导数的概念。
7 o% X' z6 z( d) U* T7 v% z
2 p: q0 A2 }" e {& l2 C" Md(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" A& T3 ` U3 @4 y8 U* P

3 t8 F! M" |8 m# ~) O% @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:4 ~& W5 u3 m1 |0 |
这样算混淆了微分和导数的概念。
0 X0 S5 `4 y+ H+ X# l' e  Y9 r+ p
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
/ _  d, u) c% U; n
) _& g3 D5 z# \, EAh, you made a mistake: it should be:+ k  r6 @9 x3 K. N3 I2 C7 h; j% W

7 V- R3 y! S' A" td(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:
: L" N( t6 T" z- Q; K0 ZYes, there should not be " ' " after "f" and "g", it is a typo. :D
$ I2 F6 Q) |. z: ^+ s d. i' }
0 @, |" f S& d5 [$ p, ^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; ~/ J: l0 Z+ s8 R7 X
# S3 Q; U6 @( k9 K8 T
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。, r7 b3 K( r1 }! S8 _
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 23:24:9 Y5 f+ [/ D4 H% p2 L! L4 ?2 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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