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:) ]6 s8 E1 T/ ?5 a1 D
请教大家一道微分题，多谢！+ T" n' s6 O- \7 I& n
9 v3 [" F5 L' T  K, s, |
d [(a+bx) ] /dx = -kc +s 6 a5 }; w: q. d! L8 _( ^, Y
where: only x and c are unknown, others are all known,  requiire c = ?
! m- }- V. {) E! v# y' t# ]3 L8 E9 @  T  [
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:
8 `# F4 U: f4 [5 x4 g7 w4 H7 m, |it is too easy:5 n5 U. X: ?/ L& B9 S( s

( d3 u- R9 l* B5 A$ Q4 wd(a+bx)/dx = b: @" k9 y# _9 B& m- j  P3 b& l

) b/ F& f  H1 {2 S( b* Pso9 Z% ^' E+ D5 R! ?
  L1 [. t3 V+ D) f
b=-kc+s
9 H+ [5 \5 i: l6 ]! j0 M+ w( a8 f/ _  V( C( o& W. p' ?  e
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:, _8 p2 B& `! [
请教大家一道微分题，多谢！
( c5 G. K+ Q! |
3 B5 L( E: o& n1 m/ A; }d [(a+bx) ] /dx = -kc +s
7 {& q% [/ x( s! A+ s: |where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?$ ~; e G. f# x' B& m
2 K% s- a1 a: g5 p
多谢了！/ l6 X" S. Z& Z- v" m. h) \
* I- q/ v$ a5 `% T! k! M
[ 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:
( M1 Q; m+ j1 q* B( w; w4 Y) Psorry, did you post another question? your statement is still not clear.
4 h4 G, e6 q, J0 s; d" m2 b/ Q0 ?6 M- k& D
a, b, s, k are constants? c means c(x) ?
# V5 D; S) ^; t# ?* [" Z' m0 g& z" L% t# W3 i' y. D
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$ d8 R# {! V8 Y& Q% U5 Z对不起，写错了，忘了变量c,应该是
2 n& v8 o3 I* _d [(a+bx) c] /dx = -kc +s " z# j2 w" M! I$ ^/ p3 k

3 O6 A2 B& Q9 a# e3 y# w: 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:
9 b4 e, n% D4 F: w: P8 C" N5 kdo you mean it is:9 q* X0 ], a% y& Z! j; D0 t! m

" X: C7 c( _  }9 }5 nd { (a+bx) C(x)}/dx = -k C(x) + s0 i( c4 _; V4 ?* ?* j7 v! H" Q

! G8 N; j& `. G+ {# D5 o  w, d7 swhere a, b, s are constants, you want to know what is C(x) ?
$ s1 E+ r1 u  d# y9 Y" U4 [6 I. q! f# y  A( s/ j$ r/ n2 T5 Q
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:
' t- h' |6 {4 {( g2 M* }& M/ z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:* h  z( P/ T7 D! E
9楼的答案有点问题吧。0 ?( m% [5 V* _' p% c
您说：1 o, F# S* H; Z3 G) z
8 ^+ {: V/ m% G- d' K* ?! |2 `
d{(a+bx)*C(x)}/dx =-k C(x) + s2 d; a/ z, d" w) [- \6 f5 I
so:; `# ?- G2 J8 L# U. v9 d# [
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
5 {* n8 G: D* |9 x3 |0 m
4 t; d7 D* h  @8 f! U, h6 b也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
: J1 p  \* c4 D2 x$ }  t9 J但这是不正确的; b& Z8 A& G+ N6 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
6 K# n A' n t: z....
7 C0 W6 [9 G. s, Xd{(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:
3 i9 \ I; A; I, Y, W: L这样算混淆了微分和导数的概念。
0 J% e0 ~/ o( h. A) X; ^9 z8 T, p! O/ P4 Q7 _
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
5 H% b1 p8 f/ G/ F+ m( y5 r# O9 d1 e
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:
" E/ v. K9 k2 ^3 N# {$ w3 @这样算混淆了微分和导数的概念。2 p! b1 N$ P$ i; n

, d: m" `& _4 I4 }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
# ~" }& X( s; h. w) d6 b8 F3 W% x2 i
Ah, you made a mistake: it should be:
9 y* {3 q e6 D
5 k1 p, ]; v% e& y d! Jd(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:
t# G [' r% qYes, there should not be " ' " after "f" and "g", it is a typo. :D0 h2 ?, r; c K& \& _

R( o' i, Y q, r" J B( JBut 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:0 H* S$ X/ d9 Z4 f. K! W- @# C; E
! }# G; P" `) g: r, h
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。
% O# Q0 q, H/ } s& }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:+ ~& J p* i9 n5 Z7 n: t: 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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