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:" l  ]4 `! @' F7 H) e3 w/ u
请教大家一道微分题，多谢！
' w. V" e* `! W1 a
7 g6 }! S3 H# b0 H1 g/ kd [(a+bx) ] /dx = -kc +s 9 C4 g2 e' z3 f- ]- A4 K& x* Z( a
where: only x and c are unknown, others are all known,  requiire c = ?5 V$ o6 N, L+ K( R
" V  Y2 x) [! R5 {" @! q; v8 O4 _$ r
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:
9 |0 l% E' f$ T- B  Dit is too easy:) @& G: i, b9 y. F7 F7 M5 \$ v) `* x

! `" X0 T# Z: S  ed(a+bx)/dx = b
2 o0 _. o/ X. e  P% K9 K! g# c) _( |. L. N. [
so6 A3 b4 C' o# Y/ _
% a, m+ W! f% C. R, A
b=-kc+s
4 T7 r8 [" q! D' Z  P9 S1 ]) J& Z& E8 K: 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:
6 D  r* ?, h4 F  \- D( S4 V( m请教大家一道微分题，多谢！/ u( l% `: h! j/ V( A
  Z1 u: e) e* k# K0 t
d [(a+bx) ] /dx = -kc +s " T, n, D. {$ O9 }, c9 {; \. l
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
# Y( ?" W) G7 O$ e% C! Z2 o9 i5 o0 x! l& [& W6 E& h1 E- }4 p, P
多谢了！6 q7 E, u$ I2 L. |; |/ c  D

/ a5 w9 a0 V' w4 ^% z! E[ 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:
2 v/ f1 E. ^, k* F. J/ P% Usorry, did you post another question? your statement is still not clear.
3 C8 _1 G6 w; B
, t. ~8 P/ D. _# ra, b, s, k are constants? c means c(x) ?
+ T) r6 ?% N: [- u6 m* }: N2 `' ]1 y; U# [- 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:
& |# s4 c7 P( U$ N对不起，写错了，忘了变量c,应该是
7 w) o H6 e' o$ m& B0 V0 ad [(a+bx) c] /dx = -kc +s & b6 b3 M( W2 _' k2 s

L, O( T# ^) y5 C# {+ P" ^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:. `4 `( ?8 y, G
do you mean it is:
& @3 i, V, E% j6 \1 R. [
5 I1 c! B) `1 T, q" Q8 ?( Td { (a+bx) C(x)}/dx = -k C(x) + s) J# k, x; ^$ |6 p

* m/ \! J8 w# b7 j% ]7 I$ B+ gwhere a, b, s are constants, you want to know what is C(x) ?* `. w Q3 I' w& x0 ~
, \) v0 Y- f+ A0 ]. s
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:# d8 q4 p; i1 V3 u
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:
* W" c5 h* i& {+ @0 j9楼的答案有点问题吧。
8 ~& w" I8 c5 O+ T% X9 T您说：9 g" d* P8 ~" G
8 S+ H% w6 P; X3 |2 K {. v1 w! G
d{(a+bx)*C(x)}/dx =-k C(x) + s
5 [" `5 I) {. v8 o. A" N( Yso:
) D/ _* D+ z/ l' \4 qbC(x) + (a+bx) dC(x)/dx = -kC(x) +s. B/ E4 b/ A6 o) _# F& X

8 t7 i/ y7 i# N1 p0 h1 \$ @+ p也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx- x5 w$ z* v9 `( Y/ m! v# p
但这是不正确的
9 U1 L! `4 U- Td{(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
* F# k& r. u% u$ P....& u/ t) M2 f' n. ?4 c* d# e; S
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:
这样算混淆了微分和导数的概念。
/ R2 \) M% A" K
. {: v+ V" c$ z9 V3 h. j- f6 c! @0 ud(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
, l; B$ N; w1 o- r( [, i* N# V' L( x; d" d6 s4 i) k) G5 V) p
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:
& {# W8 G! j7 T5 ?( E这样算混淆了微分和导数的概念。
# f- L2 C* t( j* I& S5 b- R
: A: r+ n+ J# i2 w3 d! M8 l' sd(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
) k8 e5 V. M& m/ r& _& w3 y9 E8 `1 s, E0 y1 {3 @$ n
Ah, you made a mistake: it should be:2 \6 D$ l, P$ P7 V' {
& {4 V4 y2 [: K) L
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:( v; |& {0 E2 ^2 ~6 T P i9 O
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
* t& _& w$ K5 T5 j
2 ~* z9 ~) Z QBut 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:$ ~* O$ w8 H6 j
$ \. Y' c/ Y- P$ w9 t' u6 ~% H4 V
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。- n p% O) ?0 Y8 E
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:# F" ]1 |5 L; ^0 r4 O: V
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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