moroney@jon.DEC (02/01/85)
Does anyone know the solution y=f(t) for the following differential equation?
2 ,, ,
t y (t)=K y (0)=K1, y(0)=K2
,, ,
where K, K1 and K2 are constants, y is the second derivitive, and y is the
first derivitive and t is the independant variable? This is an equation of
the motion of an object under the influence of gravity as a function of time.
Mike Moroney
..!decvax!decwrl!rhea!jon!moroneylindley@ut-ngp.UUCP (John L. Templer) (02/03/85)
> Does anyone know the solution y=f(t) for the following differential equation? > > 2 ,, , > t y (t)=K y (0)=K1, y(0)=K2 > ,, , > where K, K1 and K2 are constants, y is the second derivitive, and y is the > first derivitive and t is the independant variable? This is an equation of > the motion of an object under the influence of gravity as a function of time. > > Mike Moroney > ..!decvax!decwrl!rhea!jon!moroney Isn't this an example of a variables-seperable equation? I.e., the solution looks something like this: 2 -2 2 d t*t = K*d y Bit hard to write that on an ascii screen. Oh well, what it says is: "The second differential of t divided by t squared equals K time the second differential of y." Now just integrate twice and plug in the initial conditions. Or have I missed something? -- John L. Templer University of Texas at Austin {allegra,gatech,seismo!ut-sally,vortex}!ut-ngp!lindley "and they called it, yuppy love."
gwyn@brl-tgr.ARPA (Doug Gwyn <gwyn>) (02/03/85)
> 2 ,, , > t y (t)=K y (0)=K1, y(0)=K2 Surely this can be solved by writing y''(t) = K * t^-2 and integrating twice (using the boundary values)?
moroney@jon.DEC (02/05/85)
>Does anyone know the solution y=f(t) for the following differential equation? > > 2 ,, , > t y (t)=K y (0)=K1, y(0)=K2 > ,, , >where K, K1 and K2 are constants, y is the second derivitive, and y is the >first derivitive and t is the independant variable? This is an equation of >the motion of an object under the influence of gravity as a function of time. Sorry, folks, but I made a mistake. The equation I meant is as follows: ,, , y (t) = K/(y^2) y (0)=K1, y(0)=K2. Again, thanks in advance for your assistance. Mike Moroney ..!decvax!decwrl!rhea!jon!moroney Tue 5-Feb-1985 09:57 EST
leon@hhb.UUCP (Leon Gordon) (02/06/85)
[between meal snack]
I'm afraid that what you missed is that the original article had a
typo (or thinko??). the equation:
2
t y''(t) = K
is trivial to integrate, but it is not the equation of motion for a
particle under gravity. A point particle under the influence of
gravity (in one dimension!) is:
K
y''(t) = ----
2
y
which is considerably more difficult.
HINT: read Goldstein's Classical Mechanics (Addison-Wesley, I think)....
leon
{decvax,ihnp4,allegra}!philabs!hhb!leonlambert@boring.UUCP (02/08/85)
The problem is to integrate
y"(t) = K/(y^2).
Put v = dy/dt. Then
y" = dv/dt = (dy/dt).(dv/dy) = v.dv/dy,
so
v.dv/dy = K/y^2.
We can now separate the variables:
v.dv = (K/y^2).dy
This is easy to integrate:
(1/2)v^2 = K.(1/H - 1/y).
The H is a constant, to be determined from the initial conditions.
(H = 1/{(K1)^2/(2K)+1/K2}.) Now
dy/dt = v = sqrt(2K.(1/H-1/y).
We can again separate the variables:
dt = dy/sqrt(2K.(1/H-1/y)) = (y/sqrt(2K.(y/H-1))).dy.
By putting z = sqrt(2K.(y/H-1)) we obtain a more manageable form:
y = H.(z^2/(2K)+1),
dy = (H.z/K).dz,
so
dt = (H^2/K).(z^2/(2K)+1).dz.
This is again easy to integrate:
t-t0 = (H^2/K).(z^3/(6K)+z).
Here, t0 is a constant like H, determined by the initial conditions.
Unfortunately, this gives an implicit solution. It can be made explicit by
solving this equation of the third degree algebraically. For details on
how to do this, see the recent article <410@nbs-amrf.UUCP>.
--
Lambert Meertens
...!{seismo,philabs,decvax}!lambert@mcvax.UUCP
CWI (Centre for Mathematics and Computer Science), Amsterdam