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:
3 U$ ~2 d ~3 Q3 v( Y请教大家一道微分题，多谢！- ~+ B. D: y Y1 `& P5 Y

5 W, m$ n. }; @2 v8 v: K6 @d [(a+bx) ] /dx = -kc +s
' U3 n U1 T0 ?( mwhere: only x and c are unknown, others are all known, requiire c = ?
0 r; f5 j* g2 l3 k8 z1 [% Q( B. A) o+ 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:
, n1 ?% U2 W+ bit is too easy:5 M6 C, y, o1 H/ U' Q0 Z  Y; B8 b

" F$ Q* q0 J2 ]# ud(a+bx)/dx = b* ]8 S9 `/ p  J! i1 q' {9 ^: z
4 V" A' S& ?2 j& M9 w' A5 M
so
  U, i; b0 l: c" z( G1 _* `! Y- S' x, n5 \# m: H
b=-kc+s
! H) L9 r5 U) j" }7 \+ j. c$ M  w, k' i+ o, q$ }
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:
7 c5 J/ P+ ?' ^  ?请教大家一道微分题，多谢！3 B3 z3 a6 S  T( t
0 h: e) Z/ ~/ B8 I# s
d [(a+bx) ] /dx = -kc +s 6 z3 w% F. x6 I" l3 b
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
* `4 a5 W! J& n' W9 B- K' [1 Z/ C& D! \. w; f( \
多谢了！" g0 v" }  O% Q& X/ v1 Y

2 E8 x2 a. X7 X- `; F6 h[ 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:9 t. t1 Q2 d9 I5 F8 q. \: [
sorry, did you post another question? your statement is still not clear." U# h5 x( z  @/ `& c, ~0 L
) D+ A; `' P" J& O" K- N; y
a, b, s, k are  constants? c means c(x) ?& O$ d0 V  l, ], s' e% b
" z: Z& m- Q+ u- u  m
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:- l8 n5 u& e3 `( I: v
对不起，写错了，忘了变量c,应该是
6 X( c, N6 r1 Z4 Vd [(a+bx) c] /dx = -kc +s
/ b% Y( E- K8 Q" h# i# p
0 m; V; r4 J; f9 P3 E8 X% C4 |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:& [) A0 e& ^+ S8 d0 Y' F" M
do you mean it is:) K+ \9 }6 p2 q

; y7 X {6 S0 d6 A- L Jd { (a+bx) C(x)}/dx = -k C(x) + s7 ^$ s9 t; s- W T- [. r
0 [, @4 u' s0 Q" r# j
where a, b, s are constants, you want to know what is C(x) ?
$ a' c1 g- K0 V5 a3 d1 M1 D* j# b3 H* i
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:
- \; z4 I, s" U/ o H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:
1 N1 T1 H; D5 ^# v# s& f0 a" j9楼的答案有点问题吧。
8 s# ~" ~5 q: N z/ y1 I0 `您说：0 H5 j. P1 j, H- e. I& X$ r0 Y
" k2 C9 s" d" j' I& Z
d{(a+bx)*C(x)}/dx =-k C(x) + s2 ~4 {* E4 _& K1 K# z5 W
so:
5 h& V! v/ s7 a: |# rbC(x) + (a+bx) dC(x)/dx = -kC(x) +s. g! G6 s5 l; J+ T' `7 @3 {
- e- h% b+ m4 j8 o
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx+ i7 K; q0 g6 o5 h1 r/ V+ L
但这是不正确的
! I' V4 l/ v8 D0 }; Md{(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+ Y& c9 h4 E( a4 [3 F: F$ s1 Z
...., \4 H: A6 i5 c/ e! f O5 M
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 20:30:9 o5 d6 M$ ^. ?
这样算混淆了微分和导数的概念。
' G! M. i& j5 f1 k0 L
7 w3 G0 E" x- Y4 J( 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
( J+ P4 Q* h) D) q7 g( W2 y) C7 l( X8 s/ u, U- ^4 d# {$ r
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:
% R7 K3 |1 a, R这样算混淆了微分和导数的概念。( C1 }5 ^! T5 ~$ c

/ A+ G$ A# R0 w" yd(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
& w7 e4 ?/ g' t. T) y. N4 [5 z0 h3 a
$ L- w& m1 l: {0 X4 ZAh, you made a mistake: it should be:1 o3 _; H1 K# ^: d& z/ A
5 A' V5 F' z6 O9 M- \9 ^9 h
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:& Q; }% d+ [) u8 d3 `% l
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
2 L7 u* E, W6 @! |
+ \0 P- x" p- f$ c+ HBut 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:( m* |1 x. a6 l

* d& Z4 r0 v& {# 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。/ d& ?- I7 z' v: s( 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:! Y# V7 Q* z7 b& T( c$ J+ s, G/ a
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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