# Difference between revisions of "Science:Math Exam Resources/Courses/MATH152/April 2015/Question A 27/Solution 1"

To solve this system of differential equations, we begin by finding the eigenvalues of the coefficient matrix



These can be computed by finding the roots of 



This quadratic factors into , so the eigenvalues of the coefficient matrix are .

We now compute eigenvectors for each eigenvalue. The eigenvectors of eigenvalue 3 solve the equation 



Writing , these equations imply that , and hence . Thus  is an eigenvector of eigenvalue 3.

Similarly, for the eigenvalue 1, we have the system



so . Hence  is an eigenvector of eigenvalue 1.

The general form of the solution is therefore



Setting  and using the initial conditions, we obtain the equations



whence . Thus the solution is