Quiz 3. 4 make up solution, 3/26-3/31/03: solve the 2-dimensional system of di erential equations dy dt. 1 (cid:19). with initial condition y (0) = (cid:18) 1. Solution: first we nd the eigenvalues of the matrix: 2 t + d = 0. Since the eigenvalues are complex, we need only nd one eigenvector. Then we can take the real and imaginary parts of e tv where is an eigenvalue and v is its corresponding eigenvector to get two linearly independent solutions to the system. A linear combination of these two linearly independent solutions will give us the general solution to the system. 2y = ( 1 + 5i)x y = 1 + 5i x. An eigenvector corresponding to = 3 + 5i is (cid:18) 1. We rewrite e tv in terms of its real and imaginary parts: e tv = e( 3+5i)t(cid:18) 1. 2 cos 5t + 5 cos 5t + i sin 5t.
