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:, F, m; V" V6 L1 u3 m; \0 \
请教大家一道微分题，多谢！
5 K! V$ F' O0 ~/ h$ }3 y6 K3 `5 W; n$ C
d [(a+bx) ] /dx = -kc +s : T* e1 Y6 |; B  ^  f; H2 `( ]
where: only x and c are unknown, others are all known,  requiire c = ?
* u( u* u; M7 H2 x1 U
1 }: a  o3 c+ j+ I8 d7 n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:* g3 w9 B. Y2 ?: t& ~  C
it is too easy:
/ ]8 W7 l2 w- f' v8 `- l% U
2 K% u; e* q& d  b, K) S% Q4 H) B9 zd(a+bx)/dx = b( F1 R% p7 |" }2 A" i  n! _3 K
' k5 p/ H3 R) K
so, h, V+ G& z8 x( n
! O7 {! X% b- K( O
b=-kc+s
1 F+ J% Y1 o" c/ m' Z5 u+ S4 V, }, u( ^, g
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 20:41:9 q% M2 M: P' e( r( U$ C+ y
请教大家一道微分题，多谢！
' _  @2 k0 d6 n( m
- Y. r2 |% }' x# W* k7 _% X  {# {! Td [(a+bx) ] /dx = -kc +s ; d  d% w( T7 K& h, g# m, P3 L
where: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?' K7 V: r0 B+ f% L1 f" e& M8 }

3 {0 p% d8 o2 ?+ J1 e1 n多谢了！9 {! @  C) |. b$ l( V* j& o, ]' r1 {& }
# r8 }4 E8 f1 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:
$ ] L; A4 z" b4 ]7 U5 P0 wsorry, did you post another question? your statement is still not clear.) f+ O# `2 e# F/ o

6 c& j4 j5 B& a0 {1 R. o+ u+ U' B) G- ba, b, s, k are constants? c means c(x) ?7 x3 v0 k8 B' H" \
* Z; u$ M8 N. M$ w5 n0 K' W
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:
' `- d3 E6 K' ?5 x& i+ d U9 ^对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s , l$ A; Z: d4 b! m/ p9 n

; _5 w, H8 U# B; a& k, Y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:
- _# Q) H9 ?7 Q9 Ndo you mean it is:
. ]7 x1 M, N9 K3 Z, k2 B3 e. m
0 w7 M) e. [7 L% ud { (a+bx) C(x)}/dx = -k C(x) + s
! N J* d; ~0 b2 R* e, w
* M; p/ [1 }4 ~9 Mwhere a, b, s are constants, you want to know what is C(x) ?
' ^: f; E/ Y' c/ e/ x( I# \( {' @9 l
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:+ X3 L: N" X' f* f3 `' W0 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:! F2 H: n% @3 E6 m2 @% @3 K0 B
9楼的答案有点问题吧。8 `1 ^- A( o6 ~- R) O1 t
您说：
+ g4 `5 \2 H( w1 U" z
4 @5 q6 o" ?6 N! \9 sd{(a+bx)*C(x)}/dx =-k C(x) + s
^+ g5 `+ m7 l* }/ }5 K8 `so:
% g; S" }. b: ObC(x) + (a+bx) dC(x)/dx = -kC(x) +s
( b. g4 E" s1 b1 ^$ d! g/ B
4 S& q+ C& X' z' X" s也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
7 \- {) D# [* C) E/ Q$ G7 i但这是不正确的2 L' H- V$ m2 Y( r- 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/ L) c! X! A5 }% M( b
....! W. R2 j/ H8 v- Q4 W, l
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:
/ I( x" Y& `1 w4 s. r这样算混淆了微分和导数的概念。2 @6 k% D  @- |$ n" a9 _

4 \  y3 k3 M& B/ Bd(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
: z% c3 t$ }9 E& j' u  f
5 m8 ~' i0 [- t$ a2 c3 LIf 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:
2 x: ^  l0 ?8 q0 @) \这样算混淆了微分和导数的概念。' B& @: R* ?8 a% j1 D# v

6 O" Z8 X0 `9 }2 O  Td(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
; e  S+ j/ `8 i2 B
# f; }" _" K4 F  d8 aAh, you made a mistake: it should be:/ I5 D  C2 F3 J3 ?; O
" S' t3 o- [3 u0 \
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:
9 B3 x. z2 }- LYes, there should not be " ' " after "f" and "g", it is a typo. :D
# ~; u+ k+ ^9 k6 i/ `8 y/ k8 e# F* b R3 \& \) e2 K" n
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:6 l1 e9 z ]6 U+ V( W+ J& G# U

' D* Q* H. S) W( A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。
- o1 P+ c: q2 r( G# |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:
9 |- p$ G v" F4 D3 Q* Y1 z) x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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