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:
# o+ t! S1 r: \0 m请教大家一道微分题，多谢！" b8 m% Z$ g5 F- j6 q. P) B

( g+ o4 ~$ }% k* K2 @7 j6 dd [(a+bx) ] /dx = -kc +s
F) Y: R$ k6 n! Q6 n5 r3 V* s! O" @where: only x and c are unknown, others are all known, requiire c = ?4 x* R' f' L% j% Z5 o
0 E. t- s l5 n: t; 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:
' V7 t0 O# A: X1 v* [6 V7 `it is too easy:! y3 s5 x% C, _2 V6 l
' ^2 j$ a" Y( L) x/ v% W
d(a+bx)/dx = b
% K' ^/ N3 x9 b6 p% `3 S
& D8 }3 ~% Y3 [& U1 V# v. H+ _6 lso
* N. `4 @% F4 ?# |. h- p8 }6 R# t
5 R. s7 c6 L. Wb=-kc+s* L6 _2 x0 v$ L$ ~9 v

- O, `1 G7 h- D" H) Qfinally: 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:) J3 n( y( s. G0 h+ U
请教大家一道微分题，多谢！
2 u+ g* X6 S' @& W+ C9 K, E% t$ U5 `+ `/ X! R6 J2 \
d [(a+bx) ] /dx = -kc +s $ N/ i  x$ q: {& q% q5 F
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
/ l1 e# n" l/ R$ J& _( M
$ V+ W. M  r0 u, Q9 _$ q多谢了！
* Y0 i! x" S( z6 o3 N8 S6 u: o1 t3 E) w0 V$ \. n; y
[ 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:
' C# d) f0 ~% B9 Y, ^1 J. ?sorry, did you post another question? your statement is still not clear.
/ [+ {% A7 Q+ K9 }8 k" S$ ?( H9 m  E
a, b, s, k are  constants? c means c(x) ?
8 W% v4 P( M% e' K5 |
* ^# Q3 P  O* J2 ~! c! hplease 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:
5 j* V, j% j: @" `) Q3 F7 l( \对不起，写错了，忘了变量c,应该是8 e9 D* g1 G$ u+ F
d [(a+bx) c] /dx = -kc +s / z4 C, G- \9 M4 o; f* n
& o! ]* I) Q( s' v1 V7 J U
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:2 z; y. M- y' X9 I4 I& |4 q
do you mean it is:8 Z* f4 ]- x8 J' q4 t8 I
' V8 G$ e7 x3 o- L4 e2 M. t
d { (a+bx) C(x)}/dx = -k C(x) + s
' M6 s5 a3 e' U! C+ F2 j& j$ Y2 V: K1 O O
where a, b, s are constants, you want to know what is C(x) ?
+ p& k. c8 K+ `/ W
4 ]; Z6 r" c1 j4 G0 o! ~$ ?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 14:33:6 X) q% i( I: q' w/ T+ h# W
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:
s* C! l% g! c$ T: R9楼的答案有点问题吧。
) u# A! i2 v2 I您说：
- N, K' e5 V5 K3 {6 A
& Y) T, j* ?( h# y) j" Z) w+ {! N: Bd{(a+bx)*C(x)}/dx =-k C(x) + s5 g0 `8 Y1 g9 N7 |5 S/ o' y9 j$ ]8 E
so:& q+ N5 q: M/ q* j1 O6 M# `6 g# l
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
7 y4 T/ ~* t7 a6 C1 t5 L; @0 F5 h# g
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
. I5 L* c0 i6 \但这是不正确的
: k' r% n8 ]9 y3 e9 S+ z" N1 ld{(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$ C( h9 @$ j7 b% o+ L: s, F1 Q
....8 J2 [: M- S+ c4 ^
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 20:30:5 s/ C' g) s) A" W& `  E
这样算混淆了微分和导数的概念。
0 s/ B# e3 R" y* A' C  y
% \! k  ~8 J0 nd(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8 E" R4 n: _! \; T" w' m. E
3 E0 w" u3 y2 W1 _: p! j6 Y6 r3 I0 k; L
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:. o5 l2 G; s; Y% p2 w9 _( A
这样算混淆了微分和导数的概念。
" r8 s4 J3 D* Q3 c' R9 I) Y
5 j$ J6 z5 T1 S- B$ P" ed(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5 ~9 k; ^) e3 U" i. Y! A
4 c* Q |4 y9 c! g6 K4 M z
Ah, you made a mistake: it should be:1 A) p9 \( \6 K6 ?1 ?6 F

- i7 O& \' \$ |% dd(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:% m6 e. v g* e7 c$ X
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
1 x$ Y+ n3 E6 F l% d; M( K: i+ j, S+ ?- {8 }0 n! ~4 J
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:
- u# E/ J' V* p3 S' G. m- ]: x {0 @% 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。
! W/ _2 N/ @$ _7 @3 _4 I; G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, x7 N% R2 S( i) ]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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