数学高手请进，多谢！

醉酒当歌

请教大家一道微分题，多谢！

2 d. D2 p3 z1 b& t' b" md [(a+bx) ] /dx = -kc +s
( p; K8 _) N. l5 u( iwhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

多谢了！
7 z$ x0 \, I6 t8 N* a
[ 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:
9 [! v3 l. d& D$ a4 J' Q% i请教大家一道微分题，多谢！
2 ?- v4 _% S' `6 c
4 o7 o0 c6 a8 Y% y( V# R$ Sd [(a+bx) ] /dx = -kc +s
6 O0 s" f3 m* `8 o; hwhere: only x and c are unknown, others are all known,  requiire c = ?
4 N5 ]: `& }  d$ Q
" ^2 v" a% h. V; S. W% [9 Q多谢了！

4 @9 v0 X3 G, }1 W! {
it is too easy:
0 [' [2 [3 I' t9 W
# K2 A( g  C3 \$ Ud(a+bx)/dx = b

so

b=-kc+s

- m: q; h; a4 R, ufinally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
9 [- W2 p9 x* i2 U0 }8 S/ j9 P' uit is too easy:

d(a+bx)/dx = b

so

b=-kc+s

finally:  c= (s-b)/k

$ P2 ^  A( x8 S( b! L$ C' X
. ], g5 Q# z$ w$ nthe 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！
' e1 z% p0 N* X, [: }  R7 L
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?
! b- E3 n, `2 e1 E6 j9 ?
多谢了！
2 M; {, V# e: c9 j  h2 K2 o7 I
- g- P" R1 k: L% Z5 ?+ {[ Last edited by 醉酒 ...

, Q. L/ B5 Z2 ~) L( X5 |6 j5 i% p
sorry, did you post another question? your statement is still not clear.
% q" @9 T( W  W( C
a, b, s, k are  constants? c means c(x) ?

please 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.
5 ^8 `/ h7 m% [# j" N% s
- o9 ]4 F) I" y1 @+ o# E1 M$ e8 Da, b, s, k are  constants? c means c(x) ?

please tell me.

对不起，写错了，忘了变量c,应该是
* r4 b$ }% J* A0 x) _, F4 pd [(a+bx) c] /dx = -kc +s
: v, Y2 y8 J& @
多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
$ `& }! D* I# ^& l- H; L$ J对不起，写错了，忘了变量c,应该是
4 o, r) d7 W1 p# Wd [(a+bx) c] /dx = -kc +s
" s5 S! v# ?1 p2 \
多谢

* a5 u& I# Y2 ^& I- `  y
do you mean it is:

- E4 ]& m/ N' a' w4 ^d { (a+bx) C(x)}/dx = -k C(x) + s

where a, b, s are constants, you want to know what is C(x) ?

5 {  ?# z$ l/ d+ a  _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:
# l2 Z) ~% V9 T5 Q) \( ^do you mean it is:

+ z3 [8 ~2 b. k: J9 \9 ~d { (a+bx) C(x)}/dx = -k C(x) + s
+ D7 g. B" r0 i$ h
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 fi ...

9 k; C8 F1 [7 y- ~
, P" d; n( T; K. a5 T4 z2 G" u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:
5 c) ?- {" F2 o/ s, S  J, [; m
+ @7 F8 \( M5 a/ r0 LC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.

  x$ l1 C7 P: x
; h4 P; n2 ?6 C3 D+ n# E; Z/ }procedure:

: ]. c& v, f/ a( k. ~* zFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
* ]7 n* m5 m) {6 o: h# b- Y/ ]so:
4 h6 b1 Y/ G8 Q1 o/ t5 O
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
! l1 l( j. y/ v1 ?# H" U' ?! Gi.e.

(a+bx) dC(x)/dx  = -(k+b)C(x) +s
8 \) L# U6 T" V4 V( Z5 `+ v

) x$ }* u* z( |" F9 j9 Sintroduce 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
0 b) P8 Y0 R4 e- i% G% Ptherefore:
  e- Z6 V) b, [  w  i' r
( ^# ^1 a1 F6 c" T/ N{(a+bx)/K} dY(x)/dx=Y(x)

from here, we can get:

2 ]& N/ W* Q0 t3 K' u* x9 v) _dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)

6 C' q4 \5 R: v# ], iso that:   ln Y(x) =( K/b) ln(a+bx)

/ K* ?# c/ V- x( Zthis means:  Y(x) = (a+bx)^(K/b)
5 T/ d+ @$ e) F9 O$ pby using early transform, we can have:

, n* B, d& `2 v-(k+b)C(x)+s = (a+bx)^(k/b+1)
/ R% D: a: A8 ?  ?7 v
finally:

) x4 k! o# l+ nC(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！

: J+ f" \. M, t3 _4 @; x
9 M8 ?0 b" Y1 M2 Q; ]: {" o2 r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

9楼的答案有点问题吧。
您说：
& B: k& O8 P# w
  K- \  ~4 a6 @d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s

  F4 {1 o- f# u+ d! U; T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
6 k6 G  n) H5 G% _7 g( qd{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
7 ~0 }% i0 q* E$ u" [; E并不是   (a+bx)*C(x)   的导数除以   x   的导数。
  `) _6 D4 R, x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。
9 W0 ?* A$ H4 K# u$ R; U6 {因此：
$ d0 q; I, r6 ~) z! n6 z
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。

- H5 B. w7 ~3 s" V所以您解的有问题。

3 Q" Q% K4 q9 n[ 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:
: V6 e6 @% Z4 ~% R9 _9楼的答案有点问题吧。
您说：
( I% r, ?9 L, \0 f' q0 o5 A
. a; R$ r# w  M6 |7 O* W. Hd{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
5 r9 m8 E; D2 r" c
也就是说您认为   d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
5 S9 q8 K% b9 ?+ n4 w但这是不正确的
7 H& \- t# p4 B  ]9 Vd{(a+bx)* ...

+ f0 J5 i, j8 ?& X! S7 H  v
2 U" H, E: f% B: k- p* U3 sPlease note here:
: p3 H3 s% A( E' V* S) \7 k
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
并不是   (a+bx)*C(x)   的导数除以   x   的导数。

no "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
" R/ ~, V+ i# }- i9 {2 v! q% g0 x....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

( T3 k6 q3 }4 @) R8 p2 [
  F' X$ {2 Q, R9 _; 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
6 W4 R: w# z, _/ g
  ~8 N3 S4 ^- }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)
6 q2 D# X: d) F# k
[ 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:
这样算混淆了微分和导数的概念。
/ N6 L/ v  c' t
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
$ g. w' Y8 k# E% `. F2 L8 @
If h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

1 E8 u- Y$ X3 o* X9 J( S: M! Q
$ R$ C% a, u/ @4 h这样算混淆了微分和导数的概念。

- m) A/ i9 j1 G) z/ l3 a: x9 Q, Wd(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

8 ?( G' ]0 V4 x$ OAh, you made a mistake: it should be:
/ g1 w0 d( h; |8 y* \
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)

% }, f' q4 e- f) Z/ O8 ?5 Afor h(x)=(a+bx)*C(x)/x, you are right, But, no x there, it is:
* C% A+ H5 h7 K) U$ e3 U' ~) M7 V
/ F4 @8 j. l2 X' {/ L# D* Lh(x)=(a+bx)*C(x)

gnome-kde

Originally posted by 十年移民路 at 2005-4-6 08:45 PM:
4 g9 \  M0 H$ Z+ N' l5 H, g这样算混淆了微分和导数的概念。
+ G& V5 @) @$ n3 \
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
7 B5 J& G+ P1 K
2 Q9 |) i$ @2 ~  WAh, you made a mistake: it should be:
8 o1 u& ?8 k* d; B: o$ G
0 ~0 J* i- [: Q7 M8 Dd(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_

Originally posted by gnome-kde at 2005-4-6 20:51:
7 W0 ?& _7 X4 z, D' _" yYes, 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.

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:

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。
* [( M- h. z* A1 C2 K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 ...

& }- ]* O2 m5 c& Y# p% S4 A1 z* i+ }! R
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:
- v2 z  c) t; ~, j, g# 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？

) w' h  `( h$ y6 S+ [9 j
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。