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:' G, {! K2 b2 Z. o  e+ b
请教大家一道微分题，多谢！- j* R6 O9 R  x8 q5 j1 d

7 Q/ L/ X# A# F# L' I3 ]. x) z- xd [(a+bx) ] /dx = -kc +s
- E( Y/ A2 Q* Y* p  w: awhere: only x and c are unknown, others are all known,  requiire c = ?" S. O. {: T5 M9 ^' F1 l
3 X4 R  _2 h3 V8 s; J5 V
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:( A+ l% D5 |& i0 T3 Z/ E
it is too easy:& x6 p5 u  ~( Z6 g( T( [0 ~
  q5 }# j7 n$ w3 M
d(a+bx)/dx = b
# R. c! k1 m5 i
* K# ]! K# p- s, R; \% K7 j7 Dso
6 n+ y, Y6 g' t, \/ C6 u/ r! I/ t; z5 c& ~! H5 s  U  y
b=-kc+s
; ~# T7 B# @! w$ e4 G
4 s) B% n& j6 g, x0 _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:; `+ c5 V" v- n6 S6 d; e0 N
请教大家一道微分题，多谢！
! I& d6 f0 e9 a0 h( T( ]4 ?, r
. [2 [5 k$ v M+ r% p9 pd [(a+bx) ] /dx = -kc +s % } R2 ~! t% c9 X7 m2 E3 i5 O
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?' e o& S- R7 S0 a2 Z1 Q% z

# v1 O' I+ Y1 g1 N9 b1 I多谢了！
: U7 n0 {5 v" t, w! ?7 Y# D% [& l$ F$ f6 l# E* U! Q8 Q& o
[ 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:( s) [ L1 ]' ~. `. Z
sorry, did you post another question? your statement is still not clear.
% d0 D! X( J/ m0 O) y7 d2 V0 X; g4 W( f+ X1 |2 M
a, b, s, k are constants? c means c(x) ?" o3 n, D' v4 ?1 T; ]' X

- P6 ^1 L+ _4 G4 P# p& |# x6 U' oplease 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:) ]( h3 n0 M) ]1 R6 k
对不起，写错了，忘了变量c,应该是
* x+ e$ M6 G9 x- U* b% f& |d [(a+bx) c] /dx = -kc +s
8 Q* q( f$ G% o$ c; h3 _
+ A, F& m* K. Y( G6 p5 i# 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 v4 M& D( Z9 D3 h" K. Jdo you mean it is:( W, g" }6 _" b1 ?/ T3 F

4 A, I# A7 M8 [4 yd { (a+bx) C(x)}/dx = -k C(x) + s
; _: Q3 [" O( G
7 ]. E. C% v- o: qwhere a, b, s are constants, you want to know what is C(x) ?6 H }) q/ ^) u9 G9 R4 n9 B3 o1 _" T

- B; f9 Z$ p" x# K: e, g% Aif 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 14:33:2 F; @8 e: i) |% A2 ]
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:% \  U$ C5 d) C" E- f! h6 Z
9楼的答案有点问题吧。
! D  W: B- p+ P  y您说：4 R& v6 Q8 f0 ]
2 m* u4 }# ^( {. X5 ~
d{(a+bx)*C(x)}/dx =-k C(x) + s1 C% r( K: j  Q+ r4 J& }2 D
so:
" x  n* `! u3 |* n7 F/ W7 s* |bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
/ n* ~4 \1 j9 k
; ?; o; t4 @' R7 o也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx5 B# }, Q/ o; E% N! ]
但这是不正确的$ ^1 q# W3 R4 h& g) M& m) R# U
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: U1 T- W4 O0 z# K z/ q! y
....0 \0 J( b' u# E% M( 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:
; A7 s4 b+ _. F; A$ v) x# q$ v这样算混淆了微分和导数的概念。! n7 m! t# t8 C% ~! N+ p

2 I8 f* b" |8 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
4 T4 Q) o2 g" J) Q) B
1 X) o2 j" S# o1 H) O- }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:' q) {) \% W- I+ z' L9 h$ N
这样算混淆了微分和导数的概念。* W4 Q( @ w7 ~
+ W7 m2 K7 ?* a1 x( k6 n3 V) w
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1 w4 i( E4 `2 L9 H/ Y

$ M4 Z) H7 F [) `Ah, you made a mistake: it should be:( K; U- \5 s% {/ u- D* R1 d4 }5 s

) a5 \& x9 F! i1 Bd(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:. Y3 x+ Z' i8 S5 N. Z2 _
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
) A+ @2 |, M5 ~2 I2 z/ Z1 _% w; E) O8 n0 l4 h: P; G
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:8 T7 M% `$ W [- H5 P
4 E8 ]8 i3 x1 T3 o5 L; d/ T
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。4 | H8 a0 q% W" R; d" X# ?
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:# e3 D6 N3 V0 h3 A/ 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。

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