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:# f* L1 }$ N# L5 l
请教大家一道微分题，多谢！
! d/ R5 ^! L) _- q! q+ T9 U+ R4 h$ \' C! P
d [(a+bx) ] /dx = -kc +s
: Y! {0 |6 ]6 D! Uwhere: only x and c are unknown, others are all known, requiire c = ?5 n- ?; `$ Y$ p4 y5 `& X8 \5 I
. _" j0 |+ X+ R* s
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:: r6 U, s- O3 P% K
it is too easy:
6 k! b; H9 Z7 ?" a8 B: p, r% I# t) d2 t9 @
d(a+bx)/dx = b
5 G2 m N; O# R
% j1 _' ?9 Q# \& U) A" Oso* S" O* Z- e, Q& o6 j

' Y" ], e$ J: F* d" S: db=-kc+s
3 @9 a6 h0 L8 R+ Y% h
- |8 X! H s+ Dfinally: 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 20:41:7 r  O- r8 I% @7 U! _
请教大家一道微分题，多谢！; V+ Y9 I8 `) y0 R

+ k5 |$ M: ^( f* X  {8 Kd [(a+bx) ] /dx = -kc +s - }) l- I- T+ G  L! C3 q8 E. u. `
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
2 f5 \, @8 J' W6 ?8 s6 I
5 d0 N( t* k. t% ?2 i9 }多谢了！7 _; e! k2 y2 t4 C
7 r! |# _5 P. B7 W4 F
[ 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:
: a5 c" e* [7 {2 t; q+ J* g# Qsorry, did you post another question? your statement is still not clear.
+ y0 y0 y- p% }: t9 {
. G) N i: p7 [; O9 ?a, b, s, k are constants? c means c(x) ?9 Y2 d1 S" ^% h# l3 H2 z2 R
; u# s/ S3 k' q( N
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:
. @ c: \+ U9 ~+ ^; W ?& i3 o对不起，写错了，忘了变量c,应该是
& c5 h" K( ^9 O4 ^2 F; Md [(a+bx) c] /dx = -kc +s
' Z! R' U {" e1 u, e+ y$ R M6 @2 P% ?/ R
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 ^, T) u7 G9 h7 u
do you mean it is:
! Y5 Q% P  l8 w# G: R
1 I) t/ V$ I0 cd { (a+bx) C(x)}/dx = -k C(x) + s2 m6 D  M% T. E8 F
/ P" K, w2 A! i. j( [
where a, b, s are constants, you want to know what is C(x) ?  q- z; z& q0 {! Y0 U, ]4 H
7 M2 X( ]. ^4 u! v3 ~) J6 h
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 S% x$ J9 p( m- s1 j
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 19:34:5 H! z9 j; [  e2 ], _
9楼的答案有点问题吧。( L6 ?* m& w7 c* o- V" }
您说：+ |* G. F- |7 I0 T% E  p/ M, K( b  G  Q
3 s. L9 m" V' ]' c. O
d{(a+bx)*C(x)}/dx =-k C(x) + s) M$ N/ O; l9 g6 b
so:
! |% K+ C8 c' {5 ^bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
9 u: ?% J' V: N% }, J6 l
% ^  ^& ^: G" {% k8 M也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx, m# Q8 a" z7 j. m" g: n
但这是不正确的* d, V8 t* ]2 X* H) 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. `2 Z1 u8 t' o4 f5 H8 b6 t: N
....( J( z( D1 U% S+ K O3 l
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 _ M3 C$ z( \! l: ^( |% L* Q这样算混淆了微分和导数的概念。. Y9 Q; b& t. a2 s2 W$ o9 v
! p# F/ k' y; F0 y! \4 O$ b
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
7 R( J; r2 V( O) s3 Z
3 y+ u8 l# I6 B& n1 ?, @+ zIf 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:
6 S* F m% H+ ?& _( u这样算混淆了微分和导数的概念。
4 \; ?" b. p# K* z/ N0 J* a3 w' n: q
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* x: p" p/ ]4 p, j. F$ ^4 d) d% `3 F
Ah, you made a mistake: it should be:' Y& F9 F8 M. A, B

+ Q, L8 H6 c3 V: Q3 Zd(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:
; W" X: q. Y' K$ NYes, there should not be " ' " after "f" and "g", it is a typo. :D8 ?$ T# K1 j9 k# c' R7 o

6 M; O! K1 D. m- o$ kBut 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:
2 R# v) b8 O0 _) l& d) M1 u; M7 G) i9 T0 M% u3 D- `3 J! K3 }
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。$ t, u# y0 o' Y4 K4 f |& B% p
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:% [( S' o) E6 y' U4 U6 u
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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