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:$ h: G! ^' a! y) J$ @
请教大家一道微分题，多谢！
/ V$ M3 T2 s1 N) c" p' p/ M5 Q, Y8 T4 E. H7 ], B  i
d [(a+bx) ] /dx = -kc +s 0 ?  P0 f) ^, ?. v
where: only x and c are unknown, others are all known,  requiire c = ?
: o# s! m/ O" ~( ^, i6 N8 Q
. U) W  T+ {+ W# C. [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:
9 K* o! n* ^  j7 Pit is too easy:& @+ f$ I, S0 M! V0 r# f

. l7 S( `  K7 X) h& Gd(a+bx)/dx = b
# a8 Z  _8 m8 {7 U7 E/ {  y, K
! P* W( [& H. k8 m7 y  ?& v& a4 zso
) E9 M6 v, B% ]! @! u: C* K! R
$ V/ q2 c) v' }# Fb=-kc+s
) G4 P& X) o' d) ]& z
( F2 i$ i8 y& p) U* mfinally:  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:
, e W* y1 @7 P' m$ h4 H% f, D请教大家一道微分题，多谢！! r( q) x6 I4 |' V

# l3 h$ x' Q5 j* i% J9 X7 Rd [(a+bx) ] /dx = -kc +s * Y4 x$ ]; }" Z4 e& c" A
where: only x and c are unknown, others are all known, requiire c = function of x, what is this function?2 B# V% T; x. r* y* k* `/ L' n3 z

! }. h9 l' e$ D) _! _( |) X多谢了！$ A/ k7 ~; M/ Y$ m( o: }7 g" a
- X2 ?1 T. @; q0 F
[ 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:+ f9 k! _. z8 e
sorry, did you post another question? your statement is still not clear.
9 L$ R: n. |6 a) I9 _
: ~1 `" f0 p# T/ Y" V# da, b, s, k are constants? c means c(x) ?
- F: i5 ? c1 M2 [6 K
5 l' {" K& \# f( l2 U' cplease 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 F d( X3 h. Z* \/ ^: H7 p' R# A
对不起，写错了，忘了变量c,应该是& w; K2 \4 ?; ?2 Z6 V: F6 ]% r7 H
d [(a+bx) c] /dx = -kc +s
( i( h+ l$ ~1 R8 y7 p% f& [$ M
% C& \" u+ d7 y+ N* [3 ^4 c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:0 l1 m5 {3 w* }, h5 l
do you mean it is:
, J$ K: T% o6 w4 O5 A. q; H9 @9 x7 @, t4 `* Y: _( r, [3 A+ k% ^
d { (a+bx) C(x)}/dx = -k C(x) + s/ V3 \! T5 |' O& G

% z) p5 x/ T( }6 ?where a, b, s are constants, you want to know what is C(x) ?
& F: X4 p' N) K( ]7 e8 l1 a% S* ^& o5 \8 O
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:; s! `# r# d! z% V, j; Z4 T
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 19:34:5 m/ `1 t& }: A) I
9楼的答案有点问题吧。$ e: Z2 C1 L: e- Z  q
您说：' Q8 ~4 u, O  ~/ W% E

" g2 ]  O. {% n; [d{(a+bx)*C(x)}/dx =-k C(x) + s
( T- A/ W! V* R  l6 Vso:4 o8 F9 }9 M8 q2 ]3 ]
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s  g$ E( }+ U. c5 s; E# G0 J
( ~& g. F& _* }/ S9 S; y: S. v
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx& L1 L( j" T3 t/ D0 G4 u$ e
但这是不正确的
3 j) [3 P# s- M+ q8 i  I, fd{(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
9 U2 l$ E! {2 l$ @0 [5 m....
2 A6 i6 V3 ^6 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:
" ~: G# H G3 E% ^: b- W这样算混淆了微分和导数的概念。/ h. ?+ H( z) W; \3 [2 J! {! z
8 l( t5 Q7 P4 a- J% {2 B
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; G1 Q0 W- q4 L

! l) Y6 h n2 A% 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:
. Y; J9 m; v7 @' O这样算混淆了微分和导数的概念。1 g8 Q r7 O$ c' V: f

, F8 @5 I4 J# ]% g( m! k0 fd(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* \" O0 B; u* |8 E4 X

9 {: P) B. D( u1 {" AAh, you made a mistake: it should be:
# r3 K3 s; Z1 y4 y- P1 r l/ d: [2 }
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:
2 r7 w0 w0 m0 S% Q, ?( b9 V" m* bYes, there should not be " ' " after "f" and "g", it is a typo. :D
$ w# x+ R4 \) X, {# O r& Y& `' m0 e3 U$ r3 i
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:
9 _+ g6 j1 u: c( B" G% t! o
" m! S X9 z0 ], i$ t0 f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。
$ w9 G9 T2 B d. 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:
1 N+ D7 t" K2 P% l6 o [6 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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