Science:Math Exam Resources/Courses/MATH152/April 2017/Question B 04 (c)/Solution 1

Since the eigenvectors are linearly independent, we know that the general solution can be written as:

${\mathbf {x}}=c_{1}{\mathbf {v}}_{1}e^{\lambda _{1}t}+c_{2}{\mathbf {v}}_{2}e^{\lambda _{2}t}$

for eigenvectors ${\mathbf {v}}_{i}$ and eigenvalues $\lambda _{i}$ of the matrix $A$, with arbitrary constants $c_{i}$.

From part b), we know the eigenvalues and eigenvectors for $A$. So substituting in, we get that the general solution is:

Answer: $\color {blue}c_{1}{\begin{bmatrix}1\\{\frac {-1+i}{2}}\\\end{bmatrix}}e^{(-1+2i)t}+c_{2}{\begin{bmatrix}1\\{\frac {-1-i}{2}}\\\end{bmatrix}}e^{(-1-2i)t},{\text{ with arbitrary constants }}c_{1},c_{2}.$