# Finding general solution to system of ODEs using complex eigenvalues

Use the eigenvalue method to find the general solution to the initial value problem:

$x_1' = 3x_1-5x_2$

$x_2' = 5x_1+3x_2$

$x_1(0) = 1$ and $x_2(0) = 4$

I found complex eigenvalues $\lambda=3-5i$ and $\lambda = 3+5i$ which have corresponding eigenvectors $\left[ \begin{array}{cccc} 1\\i \end{array} \right]$ and $\left[ \begin{array}{cccc} 1\\-i \end{array} \right]$. Now I'm not sure how I can write the general solution. Does it involve both eigenvectors?

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you need to take real and imaginary parts of your solution. These two will be linearly independent real-valued solutions. –  Artem Nov 8 '12 at 0:11

If we're solving $x'(t) = A x(t)$, and $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, then $x(t) = e^{\lambda t} v$ is a solution: \begin{align*} x'(t) &= \lambda e^{\lambda t} v \\ &= e^{\lambda t} A v \\ &= A x(t). \end{align*}
Here's a different explanation. You know a basis of eigenvectors of $A$, so you can factor $A$ as $A=U \Lambda U^{-1}$, where $\Lambda$ is diagonal. (The columns of $U$ are eigenvectors of $A$, and the diagonal elements of $\Lambda$ are corresponding eigenvalues of $A$.) Now let $y(t) = U^{-1}x(t)$ and solve for $y$. The system has decoupled, and you can solve for the components of $y$ separately.
How can I ensure that my solutions are real-valued, though? $e^{(3-5i)t}$ surely isn't real. –  user1038665 Nov 7 '12 at 21:50