# Differential equation with cross product

1. Feb 21, 2007

### alsey42147

ok, i don't know what to do with something like this:

(d^2R/dt^2 ) + (dR/dt) x B = 0

where the capitals are vectors (sorry i suck at latex). R is a position vector in x-y plane and B is in the z-direction.

do i split this into equations for x and y directions separately and solve them? for x-direction i would get

d^2x/dt^2 + (dy/dt)|B| = 0

but how do i solve this? i'm pretty sure all the DEs i've ever solved had, for example, d^2x/dt^2 and dx/dt in them, but not d^2x/dt^2 and dy/dt.

or is there a quick way of doing it without having to expand the cross product?

any help would be well appreciated!!

2. Feb 21, 2007

### arildno

This is a SYSTEM of differential equations that you can rewrite as:
$$\frac{d\vec{U}}{dt}=A\vec{U}, \vec{U}=\frac{d\vec{R}}{dt}$$
and A is a matrix.
If A has constant coefficients, then the system is readily solvable with eigen-vector decomposition.

Question:
Did you get this equation from a physical problem with a Coriolis term?
Just curious..

EDIT:
Insofar as B is constant, you may gain two decoupled 3.order diff.eq's in R (i.e 2.order in U).

Last edited: Feb 21, 2007
3. Feb 21, 2007

### dextercioby

Maybe $\vec{B}$ is the magnetic field and he chose units such as q=1...

4. Feb 21, 2007

### alsey42147

damn, i wish i was more proficient with matrices and stuff. i might have to do a bit of quick revision.

the equation is just electron motion in a magnetic field without the constants.

my friend says that from the equation

d^2x/dt^2 + (dy/dt)|B| = 0

you just integrate to get

dx/dt + y|B| = constant

that doesn't seem quite right to me...is it? if it is that's all i need for now.

5. Feb 21, 2007

### J77

Is B a constant, or a function of x and y?

6. Feb 21, 2007

### HallsofIvy

Yes, writing R as <x, y, 0> you get
$$\frac{d^2x}{dt^2}+ B\frac{dy}{dt}= 0$$
and
$$\frac{d^2y}{dt^2}- B\frac{dx}{dt}= 0[/itex] You can, as Arildno suggested, introduce $u= \frac{dx}{dt}$ and $v= \frac{dy}{dt}$ and write this as a system of 4 first order differential equations. Another way to handle it is this: differentiate the first equation again to get [tex]\frac{d^3x}{dt^3}+ B\frac{d^2y}{dt^2}= 0[/itex] and use the second equation to substitute for the second derivative of y [tex]\frac{d^3x}{dt^3}+ B^2\frac{dx}{dt}= 0$$
That's easy to solve.

Once you know x(t), you can use the first equation to solve for $\frac{dy}{dt}$ and integrate once more to find y(t).

(I just noticed J77's comment. I am assuming here that B is a constant.)

Last edited by a moderator: Feb 21, 2007
7. Feb 21, 2007

### alsey42147

awesome, thanks. and yes B is constant.

8. Feb 21, 2007

### arildno

You can certainly use your friend's trick in the case of constant B. It doesn't help you an overly lot, though, compared to other techniques mentioned.