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:$ b7 m& L* a5 t, _
请教大家一道微分题，多谢！- z' P1 d8 S+ U
0 D9 M! t" f4 K* M; @1 e0 h: `" o
d [(a+bx) ] /dx = -kc +s
; E; l* w5 f# _/ s9 `where: only x and c are unknown, others are all known, requiire c = ?
( p9 U/ e6 l. G
& c, x% ~! j; W$ @/ \9 `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:  n5 B- o" p; ~
it is too easy:: X0 Q% {; \# k8 r# z
7 q+ a; q. R; I6 x! U
d(a+bx)/dx = b
6 z' d( O0 w, [
: S* i; B, ^8 w, o. W4 @so! [: ]7 Q4 s+ o* c8 i
4 T- G" h) |4 W" h3 @3 k2 a
b=-kc+s8 z4 g) H( m& Q/ i  |3 u) [$ [( u

4 L% t) j! ?5 `5 Y+ afinally:  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:+ d- W6 r/ p* o" k: G, ?& O% q% z4 \
请教大家一道微分题，多谢！
  W' Q) H9 p, x' ?* u
' @" _$ x+ E& ~8 b  md [(a+bx) ] /dx = -kc +s
/ J  K) p: x+ |. mwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?. l$ }- y' Q. C3 G2 o5 b. s1 x- B
( g& t" H! i9 ^, c2 B
多谢了！
5 F2 W5 X; D9 @/ @) `5 F! e
+ b" r" U2 X9 Z/ R[ 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:
sorry, did you post another question? your statement is still not clear.
e" j8 K3 U1 G9 `# A/ e x4 X- z7 F0 }. o! e
a, b, s, k are constants? c means c(x) ?
* l! h+ u- \" \, P; L' ^& i5 K
: n' [; _% W! splease 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 12:30:3 ]4 T2 K9 X/ b0 l0 ]
对不起，写错了，忘了变量c,应该是
& C, d. o) p, ?1 Kd [(a+bx) c] /dx = -kc +s 6 O5 |" i& ?8 x8 d3 O. y
1 b6 a" H3 F' r8 o4 O% Y c* I
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 z. {6 F" i8 \* c$ q. edo you mean it is:- d" `9 }3 v) V! k+ p& g: c

6 ~# r& H: Y" |( A# b. _1 t+ ud { (a+bx) C(x)}/dx = -k C(x) + s$ y. z" Y( y* Q0 l

% t+ e; o; j1 awhere a, b, s are constants, you want to know what is C(x) ?
9 B8 c- o) M; R: `/ Q6 Q( h# m$ d
% h* n3 a5 O; x) ]8 jif 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:
% S: v/ p# s! L, o. B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:
2 q5 m2 _1 d B7 ^9楼的答案有点问题吧。( F, F: H# `- [; _7 ^( V) n5 [2 }
您说：
0 Y3 o( S; A4 {
) v3 L/ H' Y" s7 f8 X( Jd{(a+bx)*C(x)}/dx =-k C(x) + s
& C. X$ @0 q0 ~9 I, \" C& |so:
' g. `. ?+ I/ |" O( AbC(x) + (a+bx) dC(x)/dx = -kC(x) +s+ J3 p& W- k3 J$ Y
# d( N4 J, s, z5 m! N0 X
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx7 z2 K( Z* G; B
但这是不正确的; I" c. X! w6 s8 }4 g
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
# B) T7 Q5 V1 I..... U; e) }( C0 F/ V
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:+ h* M. K2 |/ g- U7 F& a
这样算混淆了微分和导数的概念。" V. A& C# h6 j; b- l( P9 k5 `
% o' B3 ]* E7 C
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. h4 ?' s6 G) i4 K7 |

7 n9 K1 M3 N# Q% J" z) s9 eIf 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:
' ~! I1 H$ X5 @/ [, s# ]8 \( b! l, x这样算混淆了微分和导数的概念。
$ t; o4 J2 R. V. S& [! z) M
% Q! `" m5 z H! K2 J4 @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- G8 ~6 I4 |4 ~* R7 V% t

/ R2 t& O3 l" [: s& m+ Q1 @9 OAh, you made a mistake: it should be:
- U! n, E5 }+ i+ h) Z! }
% j% A( }' q9 fd(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 20:51:1 f, [$ b* ]! ^. M1 \
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
! ]7 q3 w) Q0 x% |/ q
8 w9 Y4 c S4 X9 L, g r0 f! fBut 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% j+ V& B; n8 X" ^
- j: L# ~& k+ }% d5 J8 P
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。
* p$ \1 Q! c( {+ d" w0 n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 23:24:6 _5 q) v% F/ n4 h ]7 f
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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