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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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  • $\begingroup$ you need to take real and imaginary parts of your solution. These two will be linearly independent real-valued solutions. $\endgroup$
    – Artem
    Nov 8, 2012 at 0:11

1 Answer 1

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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*}

With your two eigenvectors, you can get two solutions in this manner. Any linear combination of those two solutions is also a solution. So choose the coefficients in that linear combination to satisfy your initial conditions.

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.

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  • $\begingroup$ How can I ensure that my solutions are real-valued, though? $e^{(3-5i)t}$ surely isn't real. $\endgroup$ Nov 7, 2012 at 21:50
  • $\begingroup$ I think a uniqueness theorem guarantees that there is only one solution to your ODE (together with your two initial conditions). Also, an existence theorem tells us there is a real-valued solution. So, if you find any solution, it must be the real-valued solution (which is unique). In other words, if you just satisfy those initial conditions, your solution will turn out to be real. Let me know if I'm mistaken though. $\endgroup$
    – littleO
    Nov 7, 2012 at 22:02
  • $\begingroup$ @user1038665 Yes, since the complex eigenvalues will come in a conjugate pair, as will the eigenvector , the general solution will be real valued. See here for an example. $\endgroup$
    – Daryl
    Nov 7, 2012 at 22:12

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