In some cases, say other matrix ODE's, the eigenvalues may be complex, in which case the following step of the solving process, as well as the final form and the solution, may dramatically change. As mentioned above, this step involves finding the eigenvectors of A from the information originally provided. For each of the eigenvalues calculated we have an individual eigenvector. For the first eigenvalue, which is formula_51, we have  Simplifying the above expression by applying basic matrix multiplication rules yields  All of these calculations have been done only to obtain the last expression, which in our case is =2. Now taking some arbitrary value, presumably a small insignificant value, which is much easier to work with, for either or (in most cases it does not really matter), we substitute it into =2. Doing so produces a simple vector, which is the required eigenvector for this particular eigenvalue. In our case, we pick =2, which, in turn determines that =1 and, using the standard vector notation, our vector looks like   Performing the same operation using the second eigenvalue we calculated, which is formula_56, we obtain our second eigenvector. The process of working out this vector is not shown, but the final result is  This final step actually finds the required functions that are 'hidden' behind the derivatives given to us originally. There are two functions, because our differential equations deal with two variables. The equation which involves all the pieces of information that we have previously found has the following form:  Substituting the values of eigenvalues and eigenvectors yields Applying further simplification,  Simplifying further and writing the equations for functions formula_61 and formula_62 separately,  The above equations are, in fact, the general functions sought, but they are in their general form (with unspecified values of "A" and "B"), whilst we want to actually find their exact forms and solutions. So now we consider the problem’s given initial conditions (the problem including given initial conditions is the so-called initial value problem). Suppose we are given formula_65, which plays the role of starting point for our ordinary differential equation; application of these conditions specifies the constants, and . As we see from the formula_65 conditions, when =0, the left sides of the above equations equal 1. Thus we may construct the following system of linear equations,  Solving these equations, we find that both constants and equal 1/3.