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:. W( i5 M$ V4 U0 @& ?  o! r! a
请教大家一道微分题，多谢！
8 t+ x  O5 Z7 l3 f8 a
9 P. G# T; ?4 G/ C# O. s* Jd [(a+bx) ] /dx = -kc +s ! c& j/ f1 i  T) T5 W
where: only x and c are unknown, others are all known,  requiire c = ?
6 {6 `( e/ q% Y5 \$ e! u0 F" ]7 C" l. D+ |7 E
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:
  Z" o" y. V5 o6 u1 ]it is too easy:, B/ O: `' ~/ J* o/ y) \

& @2 R* j' S# E& G: `  ^d(a+bx)/dx = b3 w9 \3 H. V- }: [% Z2 O8 e) d2 Z
3 w+ I6 y# P5 x
so1 ^8 t9 }  o2 _) b! c
. O  S/ F0 [8 e- g) d( V. D
b=-kc+s
0 C7 N6 ~) G" @2 I
% g9 J8 U# {! N+ ufinally:  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:
5 h: L- ?' O# F0 @  K请教大家一道微分题，多谢！% \6 J3 {4 r4 D3 Y# [2 K" m
: I5 b: z  w7 L/ C6 t0 F3 L: }
d [(a+bx) ] /dx = -kc +s % B  V( `! h% d5 j5 @, ^% a
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?+ g. a5 s5 s0 O: Y
. W/ V7 C# m" M. _1 W0 M  z! Q
多谢了！' a$ e4 m' ]0 m' N& X/ \8 R! F! f) `
( b: w: X& Q. G9 M5 T! l3 v0 o0 W- ^
[ 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:" F8 Z9 V+ A# v
sorry, did you post another question? your statement is still not clear.5 S" H6 ~, q) j" q

8 X+ }' @1 }& N6 Q1 Ua, b, s, k are constants? c means c(x) ?
4 `: l# {+ c6 x! S0 K/ `" _: y; i7 q8 s9 u; ]7 {
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:
b" g1 r* r- T+ o* X对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s * ~0 F3 G W3 D$ \5 I2 N0 J

: }& @7 p* J5 Z9 w4 h9 Y. D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:
  T6 @2 x$ x- [+ C; }, Rdo you mean it is:9 f# I, z: [1 R' g
1 `5 j% ~3 m  @- u# b7 c8 R
d { (a+bx) C(x)}/dx = -k C(x) + s5 y5 ^9 T/ ~5 K% r" P0 E

, j9 D3 m6 o/ ~5 q1 Nwhere a, b, s are constants, you want to know what is C(x) ?
; X( j  p/ X9 P- {" i
" \/ Z, S9 Q7 S6 y, Hif 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:! d3 @/ f- _1 O7 j8 a! f# X8 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 19:34:1 p5 c) ^& Q8 ]) k7 W
9楼的答案有点问题吧。# t! L9 a  t6 u  F1 n6 m
您说：
/ Y  H, M0 M# y$ g5 {2 q* H+ Z3 @9 D& y
d{(a+bx)*C(x)}/dx =-k C(x) + s
5 o2 M  {, P8 d' Mso:
* P0 l, X" p+ V* K# BbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s6 `9 L7 N8 N- F

% k8 |- G$ K, v1 p  r也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
0 N; {/ }0 X2 q- k8 U- n但这是不正确的
7 s  I* O$ J# `8 V6 fd{(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$ z7 k: P2 U; |5 q$ r) _1 [$ t5 N6 z! [( c
....
0 v% e0 y; `$ Z: \( Dd{(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:( x4 T9 O% J3 k2 A
这样算混淆了微分和导数的概念。6 O$ Y9 ]( a+ a* Q1 ]& ~

. o% i& i+ N) p$ vd(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
9 O# y; |0 R! w" s
6 ?8 C2 Y- C' G: N0 F$ HIf 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:
* X7 o# J# k- l2 j这样算混淆了微分和导数的概念。
) j" W7 `1 T6 n1 m- w- M5 H. F3 p3 e6 S7 s/ h
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! O0 k- ?! s b* \1 C

+ O2 H. K7 g+ w7 ]# @% s# yAh, you made a mistake: it should be:. @* Y K/ B) C

! W7 C6 Z2 {, a5 P& \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:
0 c: ?' t! v+ L( Q$ ?3 w, X6 tYes, there should not be " ' " after "f" and "g", it is a typo. :D
/ L; |2 U/ c) G( E
. J- X; @' f; m* p, _. FBut 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:
6 z \2 Z4 n# Y- g, X+ A# k. o! h k2 k U. E7 D: q3 O8 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。
# }: C6 i6 F! v, Q' A* Z" D5 r+ Y9 F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:( v v5 G% j+ K* |! f9 S C
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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