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:
7 K" S6 @" b7 H; ~$ k; h- Y- ^, M8 G请教大家一道微分题，多谢！: Y4 Y1 ^; x% t4 O  }1 g
& x2 B; v5 q: d4 P: X" n) {7 P
d [(a+bx) ] /dx = -kc +s
' r3 j( y( i' r1 u1 o- i/ iwhere: only x and c are unknown, others are all known,  requiire c = ?2 ]" x) ^$ A- j1 X4 E- z

  Z5 l0 Y% O0 k% P+ |- 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:3 J; f: W* G8 i7 J# H# Y
it is too easy:! t9 j+ n9 k/ |& @
9 b$ V% V5 z. x7 D
d(a+bx)/dx = b; @: Q4 l- D; A5 Z  k$ u
. ^  \9 [: r0 j5 [2 R- D
so
5 C3 W6 Z! W1 }9 z9 c# i4 d2 E2 G. G+ Q4 x
b=-kc+s
& y+ Z0 @0 ]+ c# t
- @! s# @3 j# S% o: ]; G5 r: Qfinally:  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 20:41:4 J) m. N; O% V  d# ^! D3 d& g# \. ^
请教大家一道微分题，多谢！; B$ l" Q4 v  @* c; }

/ U9 C: j* P6 b( h! s- r3 V" Sd [(a+bx) ] /dx = -kc +s
% c" k, _2 }( gwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?' r& }! k7 j5 K; F7 Y
4 e* r0 M% y0 h) K! g& S
多谢了！4 I6 l0 c# u1 T* |$ f5 D* ]$ [

  e! n5 Q7 c; l5 r. x[ 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:
6 g1 _1 o) |* u2 Y+ t; j6 msorry, did you post another question? your statement is still not clear.
( x% t% ^- O6 l; e" |9 z2 d
' o6 a) }6 P" ^/ j* L2 j5 e9 g5 ta, b, s, k are constants? c means c(x) ?
M+ p, w3 y4 P% e; j+ a9 [0 M6 |' g3 s
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:" g0 H) p2 R+ j3 v, D3 H. q
对不起，写错了，忘了变量c,应该是1 K5 ` h# P5 g% e9 j/ A( f6 ~
d [(a+bx) c] /dx = -kc +s 2 `. G7 m& ?% |5 j0 d' }

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:, u3 p. @: ^8 q
do you mean it is:
7 ^2 k3 c' D( z
) _7 |; G/ ^3 z3 `2 ]d { (a+bx) C(x)}/dx = -k C(x) + s
* L2 f% ~9 {' B, v; i/ g& U; b" o
where a, b, s are constants, you want to know what is C(x) ?
" t( e* S0 J% _% v$ {0 S: @5 |4 T- o: C
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:) C# \9 G( B& l3 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:* D9 j# D: V/ p0 T! e+ x
9楼的答案有点问题吧。2 a6 P8 `, t1 j- i0 O
您说：2 V4 l' a" D4 E. _% X% W) B

3 Q! Y* Z9 P5 ]& Y8 b$ Y$ ^9 @d{(a+bx)*C(x)}/dx =-k C(x) + s
& l& I2 {' e& N* I$ M, gso:
; l) q: {# D# h2 R& W4 EbC(x) + (a+bx) dC(x)/dx = -kC(x) +s6 U& P( _+ O5 M' e
( q6 j) a% L# \) d- x
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
# C! r/ x' Z" u但这是不正确的% h% q! L5 w! w
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4 ~0 Y1 D8 ]) k
....1 ?3 _8 S* j0 U' n6 }2 }1 ?* I! s/ r
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 20:30:4 J! `! c. a3 q" g. _" _6 z8 Y
这样算混淆了微分和导数的概念。! k. M6 A% K9 b3 x: W

% X. H" H1 O8 c2 M; zd(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 C& _* ?' X# k- o: P# [4 B3 B: Z3 `2 p) o6 q
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:. m8 }% h# y' T$ l7 X! r1 d
这样算混淆了微分和导数的概念。% P4 U! |( }. _/ Y* R- J8 _% ?/ e' E

& B4 }, C9 f( A$ p, I9 b4 cd(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
5 n+ p+ T, f y7 C, k- w& Y' i/ K, I* _" H# b% r2 Q3 q
Ah, you made a mistake: it should be:
' c6 ^- v2 k ^+ K' o* b
" z$ U- f0 R, X) K) Vd(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:) e) s1 Y; u' }- ~
Yes, there should not be " ' " after "f" and "g", it is a typo. :D% H e- w1 X8 `" h. b' k

) @3 X8 ?3 m) @; z5 QBut 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:
. \! [# Q- e, R8 _, o" h
* ^- Y1 \: d ]9 [/ k d5 [ G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。& Z( W; }0 t& 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:
/ K+ {+ e$ ^$ v" k9 ^ s$ ?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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