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