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 20:41:4 M. U- k5 `# J, `& N
请教大家一道微分题，多谢！8 R. J, v; o! r

! }. j* N' V2 ~* {" j4 i% u% G$ ~d [(a+bx) ] /dx = -kc +s $ d& |5 b0 T+ B8 A; o
where: only x and c are unknown, others are all known, requiire c = ?+ y5 h: u6 Y5 g( m, d( G$ `

' P2 |' P% M& s) E1 h% t9 D9 A2 ~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:( E# Q# [0 T3 G! P- h4 L1 r+ o
it is too easy:
$ k/ z# Z( K* L4 c! {7 a/ \0 |8 y( ?- W0 I
d(a+bx)/dx = b
0 Y2 [; Y- b2 h" l1 l0 u5 F; w% V; l+ G8 e
so
7 W) t* m- y# D7 J! [
; f6 X: n! {. rb=-kc+s
9 T* W- p7 ]6 c, z1 T1 X9 S! {2 k" k' c/ R$ A) A
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:
: h( h  ~* r+ f* I请教大家一道微分题，多谢！$ V" F) Q  Y. r4 b+ W( n
. x& ]' O/ z1 U
d [(a+bx) ] /dx = -kc +s 5 w/ X0 J2 W  Q. V
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
% i3 E* R# n0 v1 S
9 r2 D* r" {1 Z# @, ]. k( k多谢了！% Y* c* ?3 E* v  ^  B; W% N8 \

+ `, p/ d# y5 F$ i7 S- \[ 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:' o' m; }8 m" }3 |* I
sorry, did you post another question? your statement is still not clear.
: F* `, W3 C) Y. k( ^4 h. A. U! E
a, b, s, k are constants? c means c(x) ?
/ C% B' d3 Y5 ?9 ?" j
) ?1 S. Z7 N: K9 _) ?: O1 l' ]8 h3 Jplease 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:; W8 g2 Y m8 @9 C
对不起，写错了，忘了变量c,应该是
+ ~6 z3 K0 J" _8 z% B9 i8 X- R# H" l0 {d [(a+bx) c] /dx = -kc +s ( g% O6 T/ f4 R8 f- ^$ N) | u
% U% v! Q: P$ B- 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:
8 l5 R6 [( P6 l) a  c% V1 K$ B) Y3 Tdo you mean it is:
0 r4 B0 r  ~" h" K3 h5 r1 q0 f; H2 a! T  `! |8 q' V; ?8 `
d { (a+bx) C(x)}/dx = -k C(x) + s
, k# q$ u* v3 h+ C
/ I) ^8 B: Z6 U+ a, W( Zwhere a, b, s are constants, you want to know what is C(x) ?
/ k1 z8 Q! S5 S( F/ ?- s4 n9 u  m. S* A
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 ^8 z F# ~# A* g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:
( B, K6 i* [9 V! u- |- P+ u9楼的答案有点问题吧。
) Q; U" h$ o3 W+ [( s* Q$ g您说：) d+ l6 L* f8 d" B2 s
" t" T7 [9 b) \8 f2 u/ b
d{(a+bx)*C(x)}/dx =-k C(x) + s
+ b, l4 k7 B$ t* g7 z7 Kso:5 B! H0 P2 M* _+ [9 k! G8 [6 n( D
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s7 `; l' ?: g; Y; A% r( G- `

f+ k3 u! P$ H0 u; V也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx% ^! L8 o0 u* Y! b& c3 W, T1 S) l
但这是不正确的
% Z5 s- {( K8 t2 j% t' ~7 Kd{(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
" `4 I4 L4 V- _* V! I....5 M: ^ i; w3 Z0 k z
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 20:30:2 C- g4 O5 M9 p4 c' _' E/ D) n
这样算混淆了微分和导数的概念。
( i9 f2 x7 t7 I# G8 i* @+ _% }' v6 U: {; d7 j
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- y) d& n1 O5 G# W' a9 `7 a4 _
6 v- r: C; ~5 V* W! t) C
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:( v$ v/ y, s. ]5 C  ?9 p- Y
这样算混淆了微分和导数的概念。
, v- e) `, v- K0 k1 m0 V7 {2 }9 D, h  J
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" }8 @6 Z7 }$ x3 w0 M) x
7 k7 e% h8 ~" p  l) p
Ah, you made a mistake: it should be:. i9 ~' N5 d& \  I$ u# j; S4 K
$ y9 T$ E& X+ S, o$ U" V# G9 A9 ?- ~
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:: i. T( x1 j* L; o" Z
Yes, there should not be " ' " after "f" and "g", it is a typo. :D6 E: R1 j) n' A+ q

B9 }3 X6 s' [7 s( _+ NBut 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:3 Y0 H8 W+ l8 l F) x* f" R8 f
; }+ @2 ?5 a; `1 _
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。0 x+ H, K# ^% L% Q8 f
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:) S9 X2 m3 _8 e1 |# Q! F, 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。

账号		自动登录	找回密码
密码			注册