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:" e3 V2 s9 a7 h7 N( x) K
请教大家一道微分题，多谢！% P; w# p6 S- U7 T# ]3 ^
% W! P7 p* g/ t8 M0 d& Y9 g5 z6 g
d [(a+bx) ] /dx = -kc +s
+ }) H. C) ^7 x. Mwhere: only x and c are unknown, others are all known, requiire c = ?
6 \- U0 v, a: q/ u! m3 R" w. m5 X+ k5 E9 J6 n& j* l
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:
" ^0 b2 g% ^/ p3 w2 V" \it is too easy:8 o) ^/ J/ U" P+ Y

! f) k  N0 p/ e, fd(a+bx)/dx = b$ v3 v( a8 }. c) M/ }% d& q

) `! [! S3 L" S( Wso4 ?+ Z; D/ y1 q: R
% C, e: Z+ F' T
b=-kc+s
3 _/ _2 V, E! [& }  h) H8 L6 B* _5 ~5 f% J& H
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_ · 发表于 2005-4-3 20:51

Originally posted by 醉酒当歌 at 2005-4-3 20:41:
# I( w% o Y3 ^7 _. \请教大家一道微分题，多谢！; K% \) y9 S) B5 p7 @. l

% x' O9 l1 x. H4 qd [(a+bx) ] /dx = -kc +s
9 t8 |) ~/ o9 ywhere: only x and c are unknown, others are all known, requiire c = function of x, what is this function?
7 H/ P: P( l! W# n# ?
5 ?/ x& D* y! S4 i+ E' v多谢了！
6 e. w0 @; @& d& a# P8 K3 M% g+ S% q9 X; B/ f3 n
[ 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:
- T6 F& X8 ~' r% Ssorry, did you post another question? your statement is still not clear.6 s8 I* x$ Y; Y9 @2 f6 S
* g& M- B  o( n( y
a, b, s, k are  constants? c means c(x) ?
- m5 R7 p+ L! C1 I1 Q9 d# Y1 t
1 j  a: `+ P! _& `) C) V' Dplease 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 12:30:5 s( p/ z! B' w4 s
对不起，写错了，忘了变量c,应该是
0 Q: ~. h5 n+ O2 B2 ^) u; p ~8 g0 ^d [(a+bx) c] /dx = -kc +s
, E. ]$ H: q ~( {- l, F7 V3 N8 L& v1 w9 {8 o; e0 M3 C T2 P
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:
4 s% l5 `4 ]! Q% d) W% Hdo you mean it is:
2 y( I) h0 w! n8 S
' L1 o' w0 t+ Q' od { (a+bx) C(x)}/dx = -k C(x) + s4 V) w! M( J, p

h+ i/ x8 C* E% G, B: Xwhere a, b, s are constants, you want to know what is C(x) ?) R8 m: a+ ?# h5 c5 {

4 h* ]7 x1 A5 l' qif 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:
5 i& m; s ~' c: J6 N7 h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:
: q& d; K- S( l8 W! p' e7 Y* j2 H9楼的答案有点问题吧。/ C: i  D$ O$ h& e! m1 o9 D2 E0 p; p
您说：- G5 N7 @- U: _) U4 _7 n
7 `8 y+ c7 L9 F3 F0 A& S
d{(a+bx)*C(x)}/dx =-k C(x) + s
! w0 b& N8 r# \' z8 r5 i7 Vso:  x0 r' m$ V0 a! S: m& J0 O
bC(x) + (a+bx) dC(x)/dx  = -kC(x) +s
& u. u$ F% B2 q( r! s4 ?" l) K+ {2 h- {1 _; V5 ]0 j$ k6 E: n; e1 g: i
也就是说您认为 d{(a+bx)*C(x)}/dx = bC(x) + (a+bx) dC(x)/dx
7 |, L. r3 M/ u1 y但这是不正确的
8 ]1 o' n: Z5 a2 j' \# \" yd{(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% A& H# s6 `3 S( X! v
....
% z/ N1 w" f" r; wd{(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: h- j( n9 W K2 O5 t+ V9 Y
这样算混淆了微分和导数的概念。! G* l: q9 R8 y% n0 j) [' e8 n

9 z# k' Z& {8 D8 \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
0 o: p; a' `9 M5 d! w" g" B; L: n1 B+ S1 P! A% A2 g8 C
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:
& ]2 h$ ~ e; }4 O; k: G- x# R这样算混淆了微分和导数的概念。
: j: ~# v! r, l/ @1 x
2 P2 |6 V: p: c" Q" u9 Ld(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 K3 u8 n; _/ j3 N u2 d; A! B7 Y9 M& k
Ah, you made a mistake: it should be:
! j! l: V; Q# ^6 @! p: {3 {
- K0 C/ t8 t2 @% \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:
: Y$ W: T6 Z+ G/ R9 LYes, there should not be " ' " after "f" and "g", it is a typo. :D
$ ]. q* C0 L3 @; L" ^
5 z3 N, `8 a" ?. u. Y2 {$ l; vBut 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:
$ C/ g2 v: p9 g$ J/ r- E
# n9 I2 D6 C7 P2 V4 H2 |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。
1 c* Z9 [ w9 H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:
* v, ?1 e8 `7 q* `! F. o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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