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:. D6 X/ \7 R. _2 D7 D" a
请教大家一道微分题，多谢！' R- r. N( ^. X# f) ^& Q
: ]7 G) `4 ]' Q+ H. v7 h# R
d [(a+bx) ] /dx = -kc +s 9 Z! \, `5 i% Z9 l' N% K/ u
where: only x and c are unknown, others are all known, requiire c = ?6 u z6 `$ l% ?0 ?

3 _: ]" z4 P b5 X5 d) D! L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:/ h: m6 ?6 S# f! {2 u+ w6 ?/ K
it is too easy:$ n% e' d; _% ]3 _( K
% F9 i: ~' b' m$ B% E
d(a+bx)/dx = b
6 x9 |5 U0 T' Q2 U& k5 M! u c
0 _& `, l7 A* I3 lso
0 z! s9 G: g0 U& i) b2 {7 K6 |+ x
b=-kc+s# z# _! p9 A4 N& X- N' P
' J+ }- @5 l' o5 x
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:
3 W8 P  q: O* @2 [: |$ N% x. v7 ~请教大家一道微分题，多谢！4 B& Y" e* Y4 z; T( h% P; ~

& h5 l$ r, _( M" S% O/ U2 w# X7 _d [(a+bx) ] /dx = -kc +s
. f& ]# w0 T' X. U, Z& Zwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?  v) Q8 f; B$ [& M( g; ~

7 S& ?: c- |# |' Q! p' Z; b多谢了！3 g2 [3 `, [( q  n
/ y9 ~2 s- {; h& P
[ 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:
& J; k& U& a+ a/ Isorry, did you post another question? your statement is still not clear.- O$ b; @6 s H3 Q e' ?
9 v4 b2 n6 U' U. x
a, b, s, k are constants? c means c(x) ?
) U& ?% o3 T* b$ J- ?5 @6 ?! ~4 c- f1 t: o( ?4 P3 F
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 L0 p3 O; e: o8 m7 U3 t+ D对不起，写错了，忘了变量c,应该是 l' r# u* g3 H; O( z: s7 X
d [(a+bx) c] /dx = -kc +s
2 @* x0 D; E- J! H5 |$ A! C! V1 |0 b6 X, j# ? E" |0 `1 m/ O
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:
& ^- b/ q0 g& a% z. Ddo you mean it is:
, v9 x: l" |% n9 v4 c0 a: @1 h* q$ E* h+ M) ?) W" y
d { (a+bx) C(x)}/dx = -k C(x) + s
: @2 z9 s1 A( B% j3 a+ Q
# f8 P z3 e$ O3 q7 Qwhere a, b, s are constants, you want to know what is C(x) ?1 Y3 I% @+ `+ w! p0 @' U
/ }) k: U) }* }3 B- o8 _
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:
$ s: Y: ?9 J5 n( x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:
9 W/ F. a. R; I5 ?) q6 Z4 b9楼的答案有点问题吧。
; E9 @7 o8 e; r- ?9 c: ^您说：- \1 B& J+ c3 u% w- G8 Q

( S. G; d; L6 r& z# ~d{(a+bx)*C(x)}/dx =-k C(x) + s
3 k" F$ l2 `$ N$ W- ?% ^! q( Dso:
! _/ E* |4 U' ^8 a$ R' B* A4 PbC(x) + (a+bx) dC(x)/dx = -kC(x) +s
; K) a6 B8 z. o& T- f
8 O8 U; a5 C. C- V也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx w, S+ B) b8 r- E! }9 E( T/ d
但这是不正确的" y0 W/ K! }: C- {) L: N
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
+ |$ d' T0 k4 g. l( d4 `! I....# v* o8 | [- M6 N
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:
- ^. u9 `, j; n& q% g& |& n这样算混淆了微分和导数的概念。 _1 F0 v' |1 y0 c& [4 |, m
1 x) @$ N8 y$ C( C& J x
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
, H( W& b: Q# E$ y
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:0 q! C' ^7 _: Q( z8 w
这样算混淆了微分和导数的概念。
, s: x' F0 U* M; q  l- G8 [& ?, W
+ {2 `# r9 L. wd(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
! y/ I  s# R6 [& i% K) l! g! K
' C% X) {8 y' ?, {% hAh, you made a mistake: it should be:
* V7 h, q7 ]6 i$ K  E" z% l$ N; Y$ _$ 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:& f, N" B! ~! h% l) L
Yes, there should not be " ' " after "f" and "g", it is a typo. :D4 m; F- V2 x: [
% z, I! Q3 X% o* V
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:( a# B' V" o: h0 c8 n

; S9 b) I9 H; F' `2 e; ?7 e, C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。4 u: m3 Y9 o6 l+ X
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 23:24:3 \& ^7 B. D( f% |1 j
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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