Science:Math Exam Resources/Courses/MATH152/April 2015/Question A 27/Solution 1

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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