# Dx/dt = y; dy/dt = x?

1. Sep 28, 2009

### Unit

x'' = y; y'' = -x?

I'm working on vector fields and I've got these two equations; no teacher in my highschool knows what to do from here.

initial conditions:
t = 0
x, y = 0, 1 (the initial position)
dx/dt = 0
dy/dt = 0

$\frac{d^2x}{dt^2} = y$
$\frac{d^2y}{dt^2} = -x$

How do I solve this?

Thanks a bundle :)
-Unit

Last edited: Sep 28, 2009
2. Sep 28, 2009

### LCKurtz

Your title gives one system and your post gives a different one. I assume your system is

x'' = y
y'' = x
plus your initial conditions. Differentiate the first equation twice:

x'''' = y'' = x, so you have:

x''''(t) - x(t) = 0.
This is a constant coefficient equation with characteristic equation:
r4 - 1 = 0.

Do you know how to write the solution for x(t) from that? If so, then you can use

y = x''

to get y. Plug in your initial conditions along the way to keep it simple.

3. Sep 28, 2009

### Unit

Yes, sorry for the discrepancy. Your assumption was right, I was indeed talking about x'' = y and y'' = -x. I fixed the thread title!

And omgsh that makes perfect sense! I was thinking I had to make a substitution of sorts, something crazy with polar coordinates or something. The fact that it's a fourth derivative equaling the first derivative makes it look like a sine/cosine function!

I'll work on the answer now, using r^4 - 1 = 0, thank you :)

4. Sep 28, 2009

### Unit

Waiiiiiit. x''''(t) = -x(t)
x''''(t) + x(t) = 0

That is what I'm solving. Does this make any difference?

5. Sep 28, 2009

### LCKurtz

Yes, a little. Characteristic equation now r4 + 1 = 0.

If you don't know how to proceed search constant coefficient differential equations on Google. Good luck.

6. Oct 3, 2009

### matematikawan

Re: x'' = y; y'' = -x?

I would rather convert the problem to the following IVP and solve

$$\frac{dZ}{dt} = AZ$$

subject to Z(0)=(0 0 1 0)t

where
$$A = \left(\begin{array}{cccc}0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ -1&0&0&0\end{array}\right)$$ $$Z = \left(\begin{array}{c}x\\ \dot{x}\\ y \\ \dot{y} \end{array}\right)$$