b*****n
发帖数: 7 | 1 y=f(x),
y'^2+2siny+cosy+1=0.
这个方程有解析解吗？万分感谢！
其中，y'是y对x的导数，y'^2 是y对x导数的平方。 |
b****t
发帖数: 22 | 2 move everything except y'^2 to the right hand side
take the square root of both sides
you get
(2siny+cosy+1)^(-1/2) dy = dx or -dx
integrate both sides
【在 b*****n 的大作中提到】
: y=f(x),
: y'^2+2siny+cosy+1=0.
: 这个方程有解析解吗？万分感谢！
: 其中，y'是y对x的导数，y'^2 是y对x导数的平方。
|
b*****n
发帖数: 7 | |
b*****n
发帖数: 7 | 4 Thank you so much for your answer!
But do you think the left side can be integrated? I cannot find the original
function, whose derivative is in such a form.
【在 b****t 的大作中提到】
: move everything except y'^2 to the right hand side
: take the square root of both sides
: you get
: (2siny+cosy+1)^(-1/2) dy = dx or -dx
: integrate both sides
|
w**a
发帖数: 1024 | 5 i don't exactly know the answer.but
i guess so, also depends on your initial y(0).
just imagine you have a pendulum, given position, it must get some time to '
swing' back and forth, under permissible condition.
【在 b*****n 的大作中提到】
: Thank you so much for your answer!
: But do you think the left side can be integrated? I cannot find the original
: function, whose derivative is in such a form.
|
w**a
发帖数: 1024 | 6 感觉像一个动力系统得东东
把y看作位置，x作为时间。
y' 就是速度。问题变成给定速度场，求位置。
有些情况方程无实数解吧，如 2siny+cos(y) > -1.
请把问题在具体一点。定义于，值域，实数解还是复数解等等。。。。
【在 b*****n 的大作中提到】
: y=f(x),
: y'^2+2siny+cosy+1=0.
: 这个方程有解析解吗？万分感谢！
: 其中，y'是y对x的导数，y'^2 是y对x导数的平方。
|
t**********r
发帖数: 256 | 7 用不着那么复杂，mathematica,maple各试一遍，如果没有解析解，
不是专业搞ODE李群分析的就可以放弃了。
【在 w**a 的大作中提到】
: 感觉像一个动力系统得东东
: 把y看作位置，x作为时间。
: y' 就是速度。问题变成给定速度场，求位置。
: 有些情况方程无实数解吧，如 2siny+cos(y) > -1.
: 请把问题在具体一点。定义于，值域，实数解还是复数解等等。。。。
|
b*****n
发帖数: 7 | 8 x确实时间，y是角位置。求实数解。
【在 w**a 的大作中提到】
: 感觉像一个动力系统得东东
: 把y看作位置，x作为时间。
: y' 就是速度。问题变成给定速度场，求位置。
: 有些情况方程无实数解吧，如 2siny+cos(y) > -1.
: 请把问题在具体一点。定义于，值域，实数解还是复数解等等。。。。
|
w**a
发帖数: 1024 | 9 你还可以对方程 求导 w.r.t. x
如果还不行， 你可以参考这本书，刚开始得几页，有类似问题。
Chaos and Integrability in Nonlinear Dynamics: An Introduction
by Michael Tabor
【在 b*****n 的大作中提到】
: y=f(x),
: y'^2+2siny+cosy+1=0.
: 这个方程有解析解吗？万分感谢！
: 其中，y'是y对x的导数，y'^2 是y对x导数的平方。
|
b*****n
发帖数: 7 | 10 Is it possible to calculate t given some y(t)=c where c is a real number?
.
【在 w**a 的大作中提到】
: 你还可以对方程 求导 w.r.t. x
: 如果还不行， 你可以参考这本书，刚开始得几页，有类似问题。
: Chaos and Integrability in Nonlinear Dynamics: An Introduction
: by Michael Tabor
|
b*****n
发帖数: 7 | 11 Still no idea.
Suppose y'(0)=10, y'^2+2siny+cosy-5=0. I made some changes to the equation in
order to make it more reasonable.
Now I want to know what the time t is when y'(t)=0.
If I cannot get the analytical solution, I am afraid there is no way to decide
the time t. Am I right?
Anyway, thank you very much, wxza.
【在 w**a 的大作中提到】
: i don't exactly know the answer.but
: i guess so, also depends on your initial y(0).
: just imagine you have a pendulum, given position, it must get some time to '
: swing' back and forth, under permissible condition.
|
w**a
发帖数: 1024 | 12 take d/dx both sides
you got
(1) y'=0. in this case there are many solutions. e.g., y=3.141592...
(2) 2y"=sin(y) - 2 cos(y). This is similar to the solution for Swinging
Pendulum.
you can see this by write all the sin and cos into one single sin() or cos().
the typical eq. governing swinging pendulum is
y" + g/L sin(y) = 0.
The perid is T= (5/4)^0.25.
as far as i know , in 2nd case, which is also the interesting one,
there is no analytical solution. if you want to study its local behavior. use
e
【在 b*****n 的大作中提到】
: x确实时间，y是角位置。求实数解。
|
b*****n
发帖数: 7 | 13 Thanks a lot!
.
【在 w**a 的大作中提到】
: take d/dx both sides
: you got
: (1) y'=0. in this case there are many solutions. e.g., y=3.141592...
: (2) 2y"=sin(y) - 2 cos(y). This is similar to the solution for Swinging
: Pendulum.
: you can see this by write all the sin and cos into one single sin() or cos().
: the typical eq. governing swinging pendulum is
: y" + g/L sin(y) = 0.
: The perid is T= (5/4)^0.25.
: as far as i know , in 2nd case, which is also the interesting one,
|
