数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！

7 e+ g) G$ t: F4 `3 ^! [' Wd [(a+bx) ] /dx = -kc +s
) S2 _7 j: w' K7 j5 r1 uwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

3 g. `& q" ?7 G) D( Y多谢了！

$ F  `2 J9 T$ E& G8 y+ H* V[ Last edited by 醉酒当歌 on 2005-4-3 at 07:56 PM ]

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-3 20:41:
: {. k$ R% T" J请教大家一道微分题，多谢！

: L) c% K" B+ P( S$ I. nd [(a+bx) ] /dx = -kc +s
where: only x and c are unknown, others are all known,  requiire c = ?

多谢了！

5 j5 \  P9 R1 o0 O' C* s
8 u6 V+ k# v+ uit is too easy:

- V0 v2 p, F. O4 Ed(a+bx)/dx = b
+ Q" ?  `. g) ~5 r  y
so
  @, g) F' @* p" F" L
+ T# \# O+ x& L6 @b=-kc+s
& ~9 j1 K+ S4 @* V. G" M
  j2 X  s1 A. i& w3 cfinally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
5 g5 F- O. V3 O  u. `it is too easy:

, h2 k8 [! v* I6 cd(a+bx)/dx = b

3 ^4 J' a- U2 a+ Z4 }3 J* E3 @so

b=-kc+s

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_

Originally posted by 醉酒当歌 at 2005-4-3 20:41:
请教大家一道微分题，多谢！

/ u+ Y8 p% ?4 W& E1 Ad [(a+bx) ] /dx = -kc +s
" ?8 N" w" ?( nwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?
; U' f) u* _6 p2 U% c8 f" T/ n- I
$ c- G- \+ z# l多谢了！

[ Last edited by 醉酒 ...

sorry, did you post another question? your statement is still not clear.

5 b9 [+ S% Z( f! h: }/ g0 q1 ga, b, s, k are  constants? c means c(x) ?

9 |1 b/ B6 e: U; j7 M! Lplease tell me.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 08:51 PM:
sorry, did you post another question? your statement is still not clear.

a, b, s, k are  constants? c means c(x) ?
# H; o2 ^2 R* p9 t/ w; L
) o: N6 i+ F7 s8 I) J7 zplease 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_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
( s% S3 W1 U* w对不起，写错了，忘了变量c,应该是
0 s8 e0 V# A! z- Q" od [(a+bx) c] /dx = -kc +s

) z' q7 e- @' c) \# k* `多谢

1 O1 o8 @' y4 L  Cdo you mean it is:

d { (a+bx) C(x)}/dx = -k C(x) + s
* N  b+ H5 D0 B
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

Originally posted by 十年移民路 at 2005-4-4 06:36 PM:
/ \6 _0 n! ^0 \" K% \8 `- B& Ddo you mean it is:
9 A5 F0 p1 \9 D4 }
d { (a+bx) C(x)}/dx = -k C(x) + s
0 }$ g% A8 T2 E. `
where a, b, s are constants, you want to know what is C(x) ?
& m5 u% |7 ^! F! q- L8 [
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_

answer:

0 z9 O* s3 j8 Y+ P* Y+ s9 vC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.

1 a+ D9 F7 n# @  |* I6 g0 g
procedure:
4 J1 b% E9 w" ]& Y! F8 ]% U/ o
( x' J; l, I3 r, D  EFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
so:

bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
i.e.
2 V1 y$ z3 V# S! _1 w( y
* y- p4 b  D: C7 c% B  {(a+bx) dC(x)/dx  = -(k+b)C(x) +s
& c; v3 q4 z8 c7 y

introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
6 k! {: [4 i) }% A4 q1 L  V% Y7 Kwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

9 E( H0 h3 E& q2 K) w{(a+bx)/K} dY(x)/dx=Y(x)

* t3 ]. y' L  d* V5 ?. V( A+ wfrom 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)
1 H- t8 K! y1 a% j9 h. ~
" s* C& e; f1 ^- ^1 Zthis 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:
* T. m- w" y$ P7 g# @" M4 {
- L( ]4 C( z; m- gC(x)= -1/(k+b)*{[(a+bx)^(k/b+1)] - s)

醉酒当歌

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_

Originally posted by 醉酒当歌 at 2005-4-5 14:33:
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！

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。
& v2 {% `% s0 L; D5 R/ U6 [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

9楼的答案有点问题吧。
您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
9 L* ^  q5 G0 V' J: U; ]# A$ Zso:
* s' Z* W$ e( q. YbC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
8 H* v2 @  u. r: }5 b/ {% ~8 f8 \
+ e' F& ~0 y0 t. F0 z  H0 V也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
! D' P( v; I- a' Y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
. M" K, F; ]0 I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
- ?$ Y/ W9 C( m! Tx' 意思是x的导数，y'意思是y的导数。分母是y的平方。
; _4 g% ~5 K( `# M6 M因此：
) ]' ^! k0 q" m: l
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。
4 b( [1 z; J. R) k6 l7 }5 m7 k
[ Last edited by 龙舞九天 on 2005-4-6 at 07:35 PM ]

十年移民路_

Originally posted by 龙舞九天 at 2005-4-6 19:34:
, O( |& K: z6 I9楼的答案有点问题吧。
3 |7 [5 y( G8 k; U! j; i% d您说：
* ]3 d" I0 ?& y& o; [! e' v
d{(a+bx)*C(x)}/dx =-k C(x) + s
9 }& B1 H2 }* Iso:
. V) t9 N4 j4 B* z. q+ |+ UbC(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
& `) G& ~4 e0 ed{(a+bx)* ...

: u* y7 T, o/ n6 q- S$ @Please note here:
8 x9 p2 ?- y% E3 B: J1 N, G4 \
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_

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
8 c- J5 m( H/ K" g4 {# M" h$ G并不是   (a+bx)*C(x)   的导数除以   x   的导数。

4 m( r  s, ?* {% N' q( s! O1 Ono "x" in that position.

gnome-kde

d{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

% e* X8 y/ j5 q
这样算混淆了微分和导数的概念。
, N9 l' W- I- L
/ M0 U" T* Q$ ?% @# ^( g/ \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 D" q3 F& H" C6 FIf 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)

3 m2 n9 ^( x3 g/ N7 d[ Last edited by gnome-kde on 2005-4-6 at 08:31 PM ]

十年移民路_

Originally posted by gnome-kde at 2005-4-6 20:30:
0 Y' t0 Q+ Z2 p' i) t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 ...

这样算混淆了微分和导数的概念。

( J# f. o, W: T% r+ R4 _1 F/ \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
2 z' r0 W4 r6 |5 ?9 c/ G
Ah, you made a mistake: it should be:

0 W$ ]6 i1 H+ f# t0 r& D3 cd(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
  ~5 H0 ?# y- q" S
; L3 ~, U' Z$ P/ ]
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)
0 I  E9 E  p" n  n  P" m+ U
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

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
这样算混淆了微分和导数的概念。
% G' D" X! y" W+ ]9 j6 |+ Q8 y* U
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

- ]6 U* |5 z) W' q% dAh, you made a mistake: it should be:

d(f(x)g(x)) ...

, @- L' S+ v+ s: `) G; d( k  q6 {
Yes, there should not be " ' " after "f" and "g", it is a typo. :D
* q8 P' S, x% U* {- g7 b
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_

Originally posted by gnome-kde at 2005-4-6 20:51:
Yes, there should not be " ' " after "f" and "g", it is a typo. :D

* Q6 K$ i7 P6 p$ U! t/ eBut for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

- k0 |6 ]  x- O
3 ]$ q" D2 W9 E1 G" u  @Thanks, I am happy here to discuss math.

醉酒当歌

Originally posted by 十年移民路 at 2005-4-5 08:41 PM:
/ [3 v: i) |; c9 O
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。
' o& c* j# ~' d- r( b  f$ W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 ...

/ l. A0 C" `. y+ P- E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_

Originally posted by 醉酒当歌 at 2005-4-6 23:24:
$ K9 W' _" l5 {2 ^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？

4 l6 |$ f1 H2 W5 @, c( j1 X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。