f*******y
发帖数: 421 | 1 y''''=(a+b*x)*y,a,b are constant. any hints or links? thanks. | C********n
发帖数: 6682 | 2 分离变量，不停地积分
【在 f*******y 的大作中提到】
: y''''=(a+b*x)*y,a,b are constant. any hints or links? thanks.
| C********n
发帖数: 6682 | 3 1 11
-y^3 ln(y) - --y^3 = (f+ex+dx^2 +cx^3 +a^x4 +x^5)
6 18
【在 f*******y 的大作中提到】
: y''''=(a+b*x)*y,a,b are constant. any hints or links? thanks.
| f*******y
发帖数: 421 | 4 多谢。分离变量后，右边积分是b*x^5+...;左边积分是不是1/6y^3ln(y)+h*x^3+...?
总共a,b,c,d,e,f,h,i,j,k常数？ | C********n
发帖数: 6682 | 5 右边只有x,左边只有y
【在 f*******y 的大作中提到】
: 多谢。分离变量后，右边积分是b*x^5+...;左边积分是不是1/6y^3ln(y)+h*x^3+...?
: 总共a,b,c,d,e,f,h,i,j,k常数？
| f*******y
发帖数: 421 | 6 敲错了，左边应当是1/6y^3ln(y)+h*y^3+i*y^2+j*y+k，不清楚11/18y^3怎么得来的？
多谢 | C********n
发帖数: 6682 | 7 逐项微分应该就看到了吧
【在 f*******y 的大作中提到】
: 敲错了，左边应当是1/6y^3ln(y)+h*y^3+i*y^2+j*y+k，不清楚11/18y^3怎么得来的？
: 多谢
| f*******y
发帖数: 421 | 8 i think the answer should be:
1/6y^3ln(y)+h*y^3+i*y^2+j*y+k=f+ex+dx^2+cx^3+1/24ax^4+1/120bx^5
The left side is 4th order derivation, so any y^3 will not have an effect.
Please let me know if i am wrong. thanks. | B********e
发帖数: 10014 | 9 咋积分 y''''/y?
【在 C********n 的大作中提到】
: 分离变量，不停地积分
| f*******y
发帖数: 421 | 10 更进一步，如果有实验数据点（x,y),如何曲线拟合上面的解？多谢。 | | | f*******y
发帖数: 421 | 11 int(1/y)=lny
int(lny)=ylny+a
...
right? | B********e
发帖数: 10014 | 12 well,
int(1/y) \neq int(y''''/y)
all i have is the formula (11) or (12) on this page
http://eqworld.ipmnet.ru/en/solutions/ode/ode-toc4.htm
(after a transformation z=ax+b of course)
【在 f*******y 的大作中提到】
: int(1/y)=lny
: int(lny)=ylny+a
: ...
: right?
| B********e
发帖数: 10014 | 13 neither
int(int(int(int(1/y))))=int(y''''/y)
or
y''''/y=ax+b <=> int(int(int(int(1/y))))=int(int(int(int(ax+b))))
【在 B********e 的大作中提到】
: well,
: int(1/y) \neq int(y''''/y)
: all i have is the formula (11) or (12) on this page
: http://eqworld.ipmnet.ru/en/solutions/ode/ode-toc4.htm
: (after a transformation z=ax+b of course)
| f*******y
发帖数: 421 | 14 my math is almost all gone and i cannot follow eq11 or 12 in your link with
substitution, especially with the 4th order, could you please explain more
why the keep intergrating method is wrong? thanks.
or if my target is to fit experimental data with a relation as the
differential equation to get estimations of constant a and b, is there any
other way to do it without solving the differential equation? it looks like
either way it will be a quite complex solution or no explicit solution at
all. | B********e
发帖数: 10014 | 15 I'm not saying Corinthian was wrong,he shouldn't have made mistake like that
. I just didn't get what he said.
On the substitution,
y''''=(ax+b)y
(want to change it to form of y''''=kxy)
simply let z=ax+b, then you view y as function of z, use chain rule:
dy/dx=dy/dz * dz/dx=dy/dz *a,
y''(x)=y''(z)*a^2 ,
...
y''''(x)=y''''(z)*a^4=zy(z) =>
y''''(z)=zy(z)/a^4
change back notation and let k=1/a^4
y''''=kxy
then you use the formula there
Numerically, I don't know. ;-(, but I don't think you should use difference
scheme, it takes too many grids.
I would rather go to the complex but explicit formula there and let computer
do the calculation.
with
like
【在 f*******y 的大作中提到】
: my math is almost all gone and i cannot follow eq11 or 12 in your link with
: substitution, especially with the 4th order, could you please explain more
: why the keep intergrating method is wrong? thanks.
: or if my target is to fit experimental data with a relation as the
: differential equation to get estimations of constant a and b, is there any
: other way to do it without solving the differential equation? it looks like
: either way it will be a quite complex solution or no explicit solution at
: all.
| B********e
发帖数: 10014 | 16 right, I would do the formula (12) with a=k, \beta=1, n=4
and let computer calculate the Mittag-Leffler function/(mainly Gamma)/
factorial
Fancy weird names draw more credits for your paper,
should be a good thing for you, right? hehe
【在 B********e 的大作中提到】
: I'm not saying Corinthian was wrong,he shouldn't have made mistake like that
: . I just didn't get what he said.
: On the substitution,
: y''''=(ax+b)y
: (want to change it to form of y''''=kxy)
: simply let z=ax+b, then you view y as function of z, use chain rule:
: dy/dx=dy/dz * dz/dx=dy/dz *a,
: y''(x)=y''(z)*a^2 ,
: ...
: y''''(x)=y''''(z)*a^4=zy(z) =>
| f*******y
发帖数: 421 | 17 could you please explain why int(int(int(int(1/y))))!=int(y''''/y)?
i use tablecurve2d to fit data and looks like no Mittag-Leffier function. as
to matlab, i donot have curvefit toolbox, can i still do it in matlab?
thanks. |
|
