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:  ~1 S1 `7 d" U% @! s8 t: L
请教大家一道微分题，多谢！, h* n1 |8 k" H
$ p  J- T" I' T% `
d [(a+bx) ] /dx = -kc +s 9 f' y- j4 S4 O3 g
where: only x and c are unknown, others are all known,  requiire c = ?' s$ {% k8 t" e9 H& ~
+ E, ~, E1 I& q% H2 Q
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:; g6 H3 ?$ ^! F% Q6 s: }
it is too easy:
# F# q# L1 W0 @7 b
4 Y* z+ B) [' _  M* Kd(a+bx)/dx = b
  n9 o: P8 z% Z: \9 m1 h/ [* a
- ^' v/ V: F% iso
2 e! |! U& [! r8 B! s9 b9 v1 E( l3 U& x. I& n/ A9 F0 T
b=-kc+s$ R; Q8 f/ H. c$ R& p( w* r
3 V( E7 g9 }2 l* A0 w
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:
+ j7 |' P; P3 N* [- I: Y" ]0 J8 Q请教大家一道微分题，多谢！
1 X8 u! a. w, j: s  B' E- X5 }3 Y2 q, f! J6 ^. s; }9 k
d [(a+bx) ] /dx = -kc +s   o9 t6 e, C4 q% X7 K" g
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?' g  ]$ |6 B7 {9 K7 E# a
2 R5 ]4 L5 M4 W- }  y9 L/ H
多谢了！
: w' L$ ]) V& [4 K6 Y- ]& w6 [! K2 X8 J1 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:
+ u, T$ W' S Rsorry, did you post another question? your statement is still not clear.% o+ Y( O0 }$ S0 M9 a
/ s; N8 S1 B5 @& U6 I( ]1 l
a, b, s, k are constants? c means c(x) ?( }# ^- Z ^ g/ K& Z

- `3 M( C. O: g' ^8 qplease 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:* k* z0 o' m4 u( F1 v6 U s M& D/ f
对不起，写错了，忘了变量c,应该是( ^' n- p& j3 u1 Q J
d [(a+bx) c] /dx = -kc +s % Z3 Z& }+ |1 W6 N, k- U

8 y7 r/ |# P1 `0 y( 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:
* C# K% S4 D1 U0 K& pdo you mean it is:
$ V) p6 p; k8 ]& {
+ h2 y0 J4 Y/ q( r5 y) B3 i ^d { (a+bx) C(x)}/dx = -k C(x) + s
9 i! B8 s: u+ u. H* q% x2 F( P O, ~1 p4 ?
where a, b, s are constants, you want to know what is C(x) ?; t8 Y: B7 s+ W
5 J3 N$ p$ @) w3 f
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:
' [& g7 w% y. Y' o) o) F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:% Y1 ?5 s# o! j
9楼的答案有点问题吧。1 G1 a& w# O8 z& w) e0 S
您说：
2 X* d, `7 [8 C6 _' H0 n& y8 ]5 [9 P$ H( N, q; S8 G* c
d{(a+bx)*C(x)}/dx =-k C(x) + s' P" S' t: _! Z5 a' W1 _
so:% u; X0 s. K2 T) f5 ~6 s% p6 C
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
0 A7 j9 C& q6 d1 \' d1 r6 E1 f, d8 @/ ?8 t
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
4 y& H+ {: _2 i3 k. ] E; ^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
) X; ]$ E) G7 W+ G, e..../ d/ _, o4 h# z) c
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:+ R( w* J8 |+ g1 H
这样算混淆了微分和导数的概念。- F' s l8 B/ F* Q
, k( }) V* P( V$ `1 T
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4 X( _# P. v; ]/ O5 z- ~' v
5 |+ N" k8 `1 V% j
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:& l+ i; C7 A2 y% ]8 l `3 [7 \
这样算混淆了微分和导数的概念。, w& Y9 ?; k% W; c( I5 N9 j; ]7 W4 Z

$ a# A5 t. E; E2 Z8 Zd(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3 Z6 u6 b% a! q1 T& O, A

; R7 m! E9 F: N) `, N0 yAh, you made a mistake: it should be:
. D% K' M7 r- Q0 q3 i* Q+ Y1 t. o8 G7 Y( B: J9 `% e4 L- i: X' `
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 20:51:5 b, y( F* @& V1 v: d
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
: q) H' y+ y5 S9 |4 v
- W3 s# h' p; k. ]# C+ \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:
! {: W& ~6 Q; F( E( q, L: \; `* r( n- C J/ g
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。
) j- P) ^) ^; J/ M8 @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 j8 N; Q( S I8 l
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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