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:
/ ^( h4 [/ {1 o, Z8 O请教大家一道微分题，多谢！
- M( j" V0 j/ }4 N7 t: Z8 }0 e1 |) J9 s* x
d [(a+bx) ] /dx = -kc +s / l. ?7 ]6 { w7 x, ?+ ]# E; _
where: only x and c are unknown, others are all known, requiire c = ?
0 l7 {/ L- X& ]3 ]+ Y
8 G; G8 u9 M& n4 p# B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:
$ Q+ M! X' u4 ?! ], U! Lit is too easy:0 t- D8 U$ H2 @2 f3 N/ t' h
: \- p( ^6 t: }, l( }5 D  t/ y
d(a+bx)/dx = b
8 [+ L/ L  R6 Z/ n# b. L1 T7 M) L( r
so4 \0 V* t, E5 E1 D

( ], I8 e; ^6 Jb=-kc+s* Z; m2 P7 h/ T, L7 s( @$ h7 _
1 c' q3 I. 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:" [. m6 {1 a/ Z% I* e0 e6 O$ o
请教大家一道微分题，多谢！: [' m5 s. X; N: r, g

4 H% W: R" H. R  ]; g# Fd [(a+bx) ] /dx = -kc +s
& k. u, P" X- n) @/ m5 T& \5 uwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?2 G, b* |) d3 I( z) ?
5 \, t; A( t  O9 C6 l5 h
多谢了！! e1 V; Q. i1 j: _% D

: ]1 T& z( _5 Z: l( f# u[ 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:0 [0 J$ z0 l4 j# \1 g
sorry, did you post another question? your statement is still not clear.3 t, u" y1 g: N0 ]( q; h

O8 ^5 F7 F/ z& U& W6 S# s( V% P& Fa, b, s, k are constants? c means c(x) ?1 W7 O3 |: t0 [ \
6 D4 I4 {! W+ K) C; [% S
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:
. b6 k# E& A, J7 a对不起，写错了，忘了变量c,应该是( B5 u6 T- y9 `& k# }+ U$ X% |: \
d [(a+bx) c] /dx = -kc +s # z/ @1 r1 w9 T6 r" r" P8 z

2 S, D+ i' j# M- e6 f6 T2 F( l y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:
9 W. p( i# z' _* Y! m+ tdo you mean it is:; h; A0 Q& W0 m5 O
5 M8 u# L5 w, u/ I: T% d# o. O
d { (a+bx) C(x)}/dx = -k C(x) + s6 D6 c. ]5 E3 t5 U! x7 w

( n3 |- i9 i4 h9 t% z& P$ _where a, b, s are constants, you want to know what is C(x) ?
7 _6 A3 J" Z4 c6 W8 F. f1 H) x# W, \
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:( q+ H# T2 q9 d0 w: N3 H& T( d
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:# @$ f: Y( `7 `4 B9 U' H4 h
9楼的答案有点问题吧。
5 q2 y' n1 g$ m! W% z您说：$ h$ q% `# @% V" M2 R5 {" p/ A
3 q/ L" @; l% y# v: t* C# n7 M2 c
d{(a+bx)*C(x)}/dx =-k C(x) + s
9 D% u& v/ F3 `so:" @3 n% g$ c8 f& f
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s' I: c! Z# b* G9 f6 s r3 |
4 M9 q% J0 Y# k) Y7 o/ K5 Y% g
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx* U6 D6 M, x% f
但这是不正确的! i. T* ?& e9 ~& Q& X, k" I
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2 p4 _; A" A) k% r6 g0 J' D& N
....% g( }' B5 g- ?: f0 V
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 20:30:8 E9 {) ?/ r% [7 z; c8 n
这样算混淆了微分和导数的概念。% i y6 r) i2 h0 A0 m; L" e0 ^$ K4 |
) r; V" {+ z X' o) D: ?, |- {
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! r0 [! I, S) C5 g6 r9 W' k: y: g

( P: N; V0 Z# @# P; U3 v# 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:
1 f  U% @7 u: T这样算混淆了微分和导数的概念。
: S: S& M3 J) l( ]: W. ~3 c& q3 u7 B9 Z6 ]$ ]9 H, t  K
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
: q) k+ t! _, ~6 ?9 i5 X! x
9 c9 y  A8 Y* D9 YAh, you made a mistake: it should be:0 n& A+ ]" R- T+ Y3 Z# e: w; ?4 n

8 |: Q- U( L# C' u4 ]$ [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 20:51:1 M0 C9 i: l( r( S
Yes, there should not be " ' " after "f" and "g", it is a typo. :D1 k/ o$ J% A- i7 M) t2 h8 j

. R+ _. z) U! K9 E. xBut 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:
& `! A9 j4 f$ R4 |1 {0 }
9 |; r7 p# Q- D2 `$ U" e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。
3 H9 j) J+ }/ \& I9 w' R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 {9 C- V$ f6 R- u; S1 E# @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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