数学高手请进，多谢！

醉酒当歌

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?
  I; O- ~- \) P# l; W
( f  k) N; K) d9 X多谢了！
! D1 e1 c- _/ U
[ 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:
3 V; w) u/ p+ v: {, O+ z请教大家一道微分题，多谢！

d [(a+bx) ] /dx = -kc +s
3 J6 q# p7 G+ H& ~where: only x and c are unknown, others are all known,  requiire c = ?

多谢了！

. z. @4 c8 g+ a5 F5 d' h
' K- e/ f6 Q8 a8 w) Ait is too easy:
4 D$ K, W1 ^9 {. u% R0 O' ?
( t2 T& X! P8 a/ Wd(a+bx)/dx = b
1 Y& ]8 _( W  {, F& T# {! Q
/ Z1 v8 S, y4 C* O, Iso
: W: |9 a  }# P% \* {
b=-kc+s

finally:  c= (s-b)/k

醉酒当歌

Originally posted by 十年移民路 at 2005-4-3 07:52 PM:
6 a  R- `5 q' R& @it is too easy:
9 B0 t! G2 t8 Y8 W2 ~
d(a+bx)/dx = b

2 A& t4 Q) c) j2 V0 l1 Yso

! d" p+ D% e8 {' ^) |, _b=-kc+s
5 n# k6 q  O( N' @6 S0 ]$ ?$ Y! z
7 n! @  K2 D' [4 p0 ffinally:  c= (s-b)/k

, d+ h( _4 m9 i7 F. c$ xthe 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！

d [(a+bx) ] /dx = -kc +s
: G8 c/ X1 f4 ?/ x) Owhere: only x and c are unknown, others are all known,  requiire c = function of x, what is this function?

: t3 K6 }1 g( ]# [9 u' [+ P多谢了！
4 r. ?. v( g: B  I9 n* c
[ Last edited by 醉酒 ...

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

' Z3 I4 v$ q6 C* l" j+ za, b, s, k are  constants? c means c(x) ?

1 z9 _' H4 e3 D% D; iplease 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.

4 U# w; M$ `- F% z4 E$ H& Na, b, s, k are  constants? c means c(x) ?

* ?: W4 N1 a, q( ^4 Qplease tell me.

! ^1 e8 D! K0 F" ^3 V对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s
6 C3 Y' e; c8 s# a$ Q
# _- X; ]) f. l* r. e- H多谢

十年移民路_

Originally posted by 醉酒当歌 at 2005-4-4 12:30:
对不起，写错了，忘了变量c,应该是
d [(a+bx) c] /dx = -kc +s

% d8 k! q, ?$ s, n2 r/ R多谢

do you mean it is:

- ^- c6 c/ e! O' C% ]d { (a+bx) C(x)}/dx = -k C(x) + s
  o: Q  Y1 |1 C2 R9 [
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:
do you mean it is:
5 A  c6 U1 K$ s. I& @- w1 R
' ?8 n" I- L7 h; ^$ Y5 Pd { (a+bx) C(x)}/dx = -k C(x) + s
0 B. n, s6 Z1 Q* s$ b
- q; A& Z( d4 E6 N) iwhere a, b, s are constants, you want to know what is C(x) ?

# c5 k: I0 P* G/ @if so, this needs to solve the differential equation. please comfirm it fi ...

+ m' I+ s+ w# F: v  ^4 K
/ H& A  o  z( {; j% D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:

, `& W7 C8 z, \* a5 |. AC(x)=-(1/k) * {(a+bx)^(-k/b) } - (s/k)

十年移民路_

this answer is the good one.

6 Z8 b) z6 @% J8 G$ ]# d
procedure:

; ^# [9 c5 p! S" YFrom:  d{(a+bx)*C(x)}/dx =-k C(x) + s
$ s* R5 T+ x9 |4 o* _so:
" f& J/ q2 ]+ D  g$ y2 ~/ K  v
bC(x) + (a+bx) dC(x)/dx = -kC(x) +s
* J# F0 K' L, C- s& i5 h3 |) O" ri.e.
" F6 x% V! X  W
(a+bx) dC(x)/dx  = -(k+b)C(x) +s

6 i7 O5 r; V3 S5 r3 n) H4 E
) a  D4 }7 Z8 }3 A$ Y3 M/ n4 }introduce a tranform: KC(x)+s =Y(x), where K=-(k+b)
* K" d$ _- C+ x( `# W4 V9 H: vwhich means: KdC(x)/dx = dY(x)/dx or dC(x)/dx = (1/K)dY/dx
therefore:

{(a+bx)/K} dY(x)/dx=Y(x)
9 e4 v$ M9 O9 ?; u& A
+ h* r) f3 B# G) Y& T* V1 Mfrom here, we can get:

/ n" f& A7 h$ E  s$ a& d) `dY(x)/Y(x) = [K/(a+bx)]dx  i.e. dY(x)/Y(x) = {K/b(a+bx)}d(a+bx)
2 d- o# S7 F/ u6 Y4 E
so that:   ln Y(x) =( K/b) ln(a+bx)

0 t9 U# e/ L7 t8 j* U; a$ bthis means:  Y(x) = (a+bx)^(K/b)
by using early transform, we can have:

3 _8 B" x5 r4 U( U-(k+b)C(x)+s = (a+bx)^(k/b+1)
" U4 g2 ~) }8 e. \2 f  W7 q' W
* i- {7 S) @  I: efinally:

C(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:
" C# V( B! M4 d' L- V6 n/ N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

9楼的答案有点问题吧。
您说：

: }/ }! U9 E2 G; j. X, y! ^' Wd{(a+bx)*C(x)}/dx =-k C(x) + s
so:
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
) C0 k) H# {' i# R3 D
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
3 b' O9 N0 _) i7 o) bd{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
! Z8 a2 I. L  ?  v2 P# n& W$ F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
7 G* h' H  l0 n6 e! n5 |" D8 fx' 意思是x的导数，y'意思是y的导数。分母是y的平方。
) S" M3 E  t& `3 Y1 Q( f: \  D+ w1 F因此：

5 ]! l/ @5 H+ w1 Vd{(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。
& T7 z# P) H/ I1 B" _$ q
所以您解的有问题。
% F/ a! ^) y1 y. _% [  [2 {
/ P/ q0 b  Z" b' b5 M[ 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:
9楼的答案有点问题吧。
您说：

d{(a+bx)*C(x)}/dx =-k C(x) + s
so:
6 F0 y- e' ^  f1 f) ?( ObC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
! d9 `( l* t! Q, f
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
/ }" k, i8 z  V; u1 V) G: fd{(a+bx)* ...

  m2 ]' m% m' X
, _8 c7 U) B( @Please note here:

1 i. N( X7 w1 U+ w; ?d{(a+bx)*C(x)}/dx  means:   df(x)/dx, where f(x) = (a+bx)*C(x)
' g+ N$ e* V; ?9 E0 u4 i( D" l8 ait is a multiple, not division. Right ?

十年移民路_

但这是不正确的
, {) N# L8 e, @2 I) o$ t1 n) Vd{(a+bx)*C(x)}/dx  意思是说对   (a+bx)*C(x)/x   整个代数式求导数
: W# l$ O  y5 B- u0 M9 T并不是   (a+bx)*C(x)   的导数除以   x   的导数。

' ^4 l& W6 K" E8 w) M3 n8 g% `' xno "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; K, e, a3 _& u- X: D5 G: o....
d{(a+bx)*C(x)}/dx =[xb*c(x)+x(a+bx)*c'(x)-(a+bx)*c(x)]/x2

: ~4 q) @: J. ~9 I- a: ~
% W+ ]! x' Q8 L5 F' e- V0 }9 |这样算混淆了微分和导数的概念。
! J  ~. L9 |4 j2 z- x" N; |+ [' H
/ I0 p; w, C1 ?5 Q  `0 Bd(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)
( b, v4 ?, a# K$ t6 k% f
[ 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:
( d+ U2 v1 s8 ~4 F- [8 j, J% n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

# t  b1 V8 y, jIf h(x)=(a+bx)*C(x)/x, then d(h(x))/dx = d{(a+bx)*C(x)/x}/dx = h ...

. M0 z( b# {0 }
这样算混淆了微分和导数的概念。
, c9 X: H% p0 _, Q! M
" ?( i9 H& Z9 ?4 p4 {1 K1 Kd(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
+ m8 I# E" Y8 I* J
Ah, you made a mistake: it should be:
# ?3 j5 X; o- @9 \. N& a( b
d(f(x)g(x))/dx = f(x) d(g(x))/dx + g(x) d(f(x))/dx
0 H6 x/ c3 ~( s& N! @

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:
3 G: b$ u" m% H8 d) d5 B* D  t
" P! \' n9 u0 s1 ~5 `: S* \/ z& x0 Ch(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。

8 \' b& k! `4 u: Z) `5 l& _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

) M, Y1 T9 S4 o, P4 WAh, you made a mistake: it should be:
, q% K8 j4 c) ^1 @% C0 d; ^
7 `0 i1 I3 e8 w, \- w; }; f5 fd(f(x)g(x)) ...

4 b7 F" [* }7 O' R3 j& U& fYes, 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:
4 l. f% Z5 e" b. a9 ?1 uYes, there should not be " ' " after "f" and "g", it is a typo. :D

# I9 L" K* w  }$ q9 W! qBut for the left part, I am not intented to show you are wrong but "龙舞九天" is wrong.

" i2 x; }9 P6 a! ~+ IThanks, 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:
1 r" x; V/ B/ C0 h! a. F* ^7 V: h# F
: n" \. u: v- G  V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 ...

# L3 B! d; N: S- y
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:
5 D5 Y  h, D% v- i3 d) L+ j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？

6 b  c: F5 Y2 j  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。