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:
0 x1 e8 }! a; U/ y  W" Y: ~请教大家一道微分题，多谢！
8 f' C  i( l3 [' ~9 G- t+ _! ?8 o
2 F; S  v+ ]6 D1 g" I" F5 wd [(a+bx) ] /dx = -kc +s
0 J/ p% a5 A! f4 N; H6 o9 ^# I: Rwhere: only x and c are unknown, others are all known,  requiire c = ?
. o' }  y% u! g# G1 G( M* q4 j
- h; C9 _: c! o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:
$ I- |" T% N( Pit is too easy:
6 P  j6 b3 {5 C4 E3 A" Q! z% V) O% w# w7 I% z; P& ?0 ]
d(a+bx)/dx = b
. a1 ]/ V- f3 v& t$ r0 P1 |, J$ t& A# s' v- P4 s: y6 K
so, ~3 C4 v7 Y" r) Z0 P9 L

6 h% s5 M" p" P# k' M  _b=-kc+s
/ K/ D5 `4 D8 v, w( R8 D- I; q8 q( r% \; V6 W& w7 p8 _
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:
2 ?/ k$ j7 o" `& E3 h1 v请教大家一道微分题，多谢！+ _$ S% |0 J$ k
! V( e0 n4 R: L0 W7 a0 m
d [(a+bx) ] /dx = -kc +s
& R0 T7 D" b1 Uwhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?$ M2 {! a* o$ k4 R, C
, C3 z% `/ d# j0 q$ w
多谢了！$ v; b5 \6 |. v) U0 Y* v: Y: {

+ r# A4 @6 a! o. S+ H" c[ 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:6 @ J: V/ A7 F( a4 Y
sorry, did you post another question? your statement is still not clear.
6 l1 o: j0 R, L9 R7 q0 Z( a+ q
5 Y. s% }7 `! P. X. ka, b, s, k are constants? c means c(x) ?' g3 b' D* ~1 L, z% q
0 K. G; D5 z/ U. R
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:
; A( g5 `8 T( C. M, U, y7 n0 L" S对不起，写错了，忘了变量c,应该是
7 X2 `0 U. m; p* S8 \+ n2 bd [(a+bx) c] /dx = -kc +s
9 |4 P, C5 J2 ^# j
1 \& a7 N" R' _4 S% 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:% [: L# S* F" b! D) |0 `3 E
do you mean it is:! K$ i1 {0 ^0 ]2 M
1 z- L# I+ ^# k- U( r9 [% b( R% O
d { (a+bx) C(x)}/dx = -k C(x) + s0 p+ X5 b" b) }) l* _1 C4 y6 q

2 I' Z4 j" R+ n/ Z. t7 Rwhere a, b, s are constants, you want to know what is C(x) ?) @. K8 S5 q0 W+ ]9 o! ^1 @

& V! c: }9 I" J( |1 Kif 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 14:33:7 o2 m% m# e2 C$ U' a8 @
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 19:34:4 b9 R+ G: T, O% [* X
9楼的答案有点问题吧。# v* D% M1 Y/ t% L4 u2 q
您说：
* {2 S5 A; |4 @( w' F0 E! X! ]* X7 B7 o; l+ G
d{(a+bx)*C(x)}/dx =-k C(x) + s
" b: J# K8 c- b5 p& l0 Tso:2 p; ^; o' k, L( R2 m
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s! d; C7 K, q" l: j
% e% P3 H: C: f7 k7 n% ?
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
6 J7 m6 J1 M3 y7 ?+ L- ]但这是不正确的
* ]0 V: y; m5 c2 A$ u% _) z& \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
" P+ L/ s! l! v M/ ]& P$ e* k1 U....
0 q/ ?9 D" E5 u. |0 g3 Z6 }$ kd{(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:
& j) I; ]8 o- K. c5 z) D* R7 G这样算混淆了微分和导数的概念。- i/ |) T+ Z" o0 i. z# ^5 E% @. }

9 b k. M' p- x$ V* 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) E% Q& p" E1 _* R

# K: B' [+ J8 m# ~0 X( bIf 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: |8 t+ S/ M8 ^1 `- Q
这样算混淆了微分和导数的概念。
' K, n+ M' \- Y# X6 g8 O3 ?) M7 j7 P0 F% B+ Z) Q% t c* ~* m
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
+ I+ @- E) B5 a0 V! W6 \$ k3 z9 }8 P3 h
Ah, you made a mistake: it should be:
+ _+ U3 p7 t. W$ o. o7 R9 {/ e; A( ]: b8 S5 q
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:
! o5 k8 X0 l4 N# l2 K) \Yes, there should not be " ' " after "f" and "g", it is a typo. :D
! Z, W4 q; w! u) A' E* _
+ D) [& u% \" i/ yBut 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:
. P% ^6 d& R& E$ b4 ~2 |- z3 V" e# w$ G( K, Q. h, }; q4 M
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。# \& b- w% p0 ~) R3 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:
: i( Z7 B/ C# M) I7 m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。

账号		自动登录	找回密码
密码			注册

数学高手请进，多谢！

浏览过的版块