Take a look here:

http://farside.ph.utexas.edu/teaching/336k/Newtonhtml/node56.html#e8.8

Demetris it is definely funny that a problem originating from Fitzpatrick, Plasma Physics, chp. 2 comes back to Fitzpatrick, especially because I posed the problem in an analytical form such that the physical origin was not even cited, but also hidden I think. Or maybe this kind of equation arises only in guiding center motion? Anyway, I am reading carefully Fitzpatrick the 2nd... :-) Thank you!

a system means there are many independent variables so you must transform your  problem to a system,then solve it.

Matteo, Adil is right. There is nothing special about a vector equation, except that the equation takes into account spatial dependencies. If the system is isotropic, then the equation reduces to a scalar equation or to what you are more familiar seeing when reading and solving differential equations.

You essentially need to take into account each orthogonal direction, the dependencies that might exist along those lines and then separate the equation into its several scalar differential equations.  A great example of this is Maxwell's equations for electromagnetism. These equations are fundamentally vectorial in nature, but most of the solutions we see in textbooks are in scalar form, because most authors assume isotropic spaces.

If you cannot reduce the pde to a scalar space, then Maxwell's equations break out into three equations, each representing the behavior and dependencies along each spatial direction. If the problem has dependencies that do not allow separation, that is to say, the coefficients are interdependent along two or more lines, then you have a far more complex problem - termed as not being linearly dependent, that is, the system is linearly interdependent.

Now, in some rare cases, you can still find a solution for linearly interdependent systems, usually after much work on your part. Worse than this would be coefficients that are also time dependent, then the problem becomes truly intractable, where you must break up the solution space to account for changes in time and space! 

So, the general advise is to try to reduce the equation as much as possible, by assuming or simplifying the problem in degrees of freedom. The simpler equation may not be exactly representative of "reality" - most solutions do not take into account all that is going on in reality (as best we understand, of course) - nevertheless, the solution to the simpler system of equations may capture the bulk phenomena occurring within the real world dynamic system. This will provide insight and ideas for further complication. 

If $\vec w=0$ there in no differential equation. If $\vec g=0$ then $r(t)=\alpha (t) \vec w$ and $\alpha$ free.

If $\vec g\neq 0$ then necesarily $(\vec g,\vec w)=0$,

otherwise there is no solution. So $\vec g$ and $\vec w$ are not collinear and ortogonal, so taking $\vec e=\vec g \times \vec w$ then $\vec g$, $\vec w$ and $\vec e$ is a basis. We propose $r(t)=\alpha(t)\vec w+\beta(t) \vec g+\gamma(t) \vec e$, obtaining

$$

-\vert w\vert^2\gamma '(t)=1, \beta'(t)=0

$$

and $\alpha$ free.

 Dear Matteo

Just represent the cross product as a matrix product i.e form the matrix of the linear 

map defined by the crossproduct and you get a set of constant coeff. ODE:s. 

So the left hand side is \vec \omega \times \frac {d}{dt} \vec r 

and equation can be represented with a matrix product as  A\frac {d}{dt} \vec r= g 

If  omega=(w1,w2,w3) then

the matrix A of a crossproduct is 

A=[[0,-w3,w2],[w3,0,-w1],[-w2,w1,0]]

where inner brackets are rows of the matrix. 

The crossproduct has of course a matrix since its linear

omega X (ax+by)=a*omega X (x)+b*omega X (y)

Now the system is in standard form 

Ar'(t)=g

Note that the matrix of a crossproduct is not of full rank so there will be a zero eigenvalue which correspond to the vector which has the same direction as omega=(w1,w2,w3) other eigenvector (or eigenspace to be more exact) has algebraic multiplicity 2 and the eigenspace corresponding to it is of course the plane which is perpendicular to omega=(w1,w2,w3). 

So homogeneous system Ar'(t)=0 will have a constant solution corresponding to eigenvalue 0 and to eigenspace whose direction is omega.

The 2 solutions correponding to multiple eigenvalue (|omega|^2 ) are in the plane perpendicular to omega and are 

proportional to exp(t)v1 and texp(t)v2 where v1 and v2 are vectors from the plane perpendicular to omega. 

The general solution to homogeneous eq is then 

r_h=c*omega + exp(k1t)v1+texp(k2t)v2 

and the complete solution is r_h+r_s where r_s is the special solution to system. If you wwould send me a pdf with actual coefficient I could solve this for you quickly. 

But I hope this helps ! 

kind regards

Samuli Piipponen

UN of Tamperre, Finland

I think the solution by Samuli of converting the cross product into a matrix product is the most general and elegant one among all, and it is in the exact form I was looking for. In the following, I try to reformulate and explain the solution steps, and I add a more general solution in tensorial form.

The approach is to reduce one complicated problem to many simple ones of which the soluton is known, and the to relate them and recollect their single solution to build the general solution of the original problem.

First, the fundaental idea is: if only we could transform the cross product operator into a matrix operator, i.e. convert the vector equation in three dimensions:

k ^ x = b

(where k, b are constant vectors, x is the variable vector, and ^ is the cross product) simply into a particular linear system:

A x = b

where A is a matrix and x and b are vectors, then the solution of this linear system would be very simple:

x = A^-1 b

where A^-1 is the inverse matrix of A. Then our problem becomes to find a matrix that operates on the x vector exactly in the same way of the cross product by k, so a solution of the equation:

A x = k^ x

If we consider the cartesian coordinates:

x=(x1, x2, x3)

k=(k1, k2, k3)

the matrix A corresponding to the k ^ operator in cartesian coordinates:

A(k1, k2, k3) = ( 0 -k3 k2

k3 0 -k1

k2 k1 0 )

In other coordinate systems (i.e., cylindrical or spherical), the matrix representation of the second-order tensor A becomes obviously different, but one can obtain the new matrix, the one associated ot the new coordinate system, by multiplying the cartesian matrix A(k1, k2, k3) by a transformation matrix converting the three cartesian coordinates into the three new ones.

If one switches to the tensor notation, the original equation is in tensorial notation written as:

e_ijk k_k x_i= b_j

where e_ijk is the Levi Civita symbol. The second-order tensor A is the solution of the equation:

A_jk = e_ijk k_k

and we can call it a "zoom + rotation matrix", since it rotates the vector x_i and it scales the vector of a factor k sin alpha, where alpha is the angle between the vector k and the vector x.

Actually, the original problem I posed was actually slightly different, since it involved a first-order derivative of the variable:

k(t) ^ x'(t) = b (t)

Where x'(t) is the derivative of x with respect to the variable t, and k and b are also depending on this variable, one could convert it into the linear system excatly in the same way as before:

A(t) x'(t) = b (t)

which here becomes a linear O.D.E. system with variable coefficients, whose solution is:

x(t) = \int_t0^t A^-1(tau) b(tau) dtau + x_0

where x_0 is the initial value fo the variable (the value for t=t0) and int_t0^t ... dt' is the integral from t0 to t in the variable tau.

The tensor A is excatly in the same as in the non differential problem.

So finally thank you very much Samuli!