I want to solve the following differential equation using eigenvectors:


$y' = 2x-5y$

Or using matrices: $\begin{pmatrix} x' \\ y' \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ 2 & -5 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}$

With the initial conditions $x(0) = 0$ and $y(0) = 2$

Now I know that all I'd have to do is essentially find the exponential of the matrix
$$ A =\begin{pmatrix}0 & 1 \\2 & -5 \end{pmatrix}$$ but my teacher wants us to solve it using eigenvectors.

Let's find A's eigenvectors:

$det(A-\lambda I) = -\lambda(-5-\lambda)-2 = \lambda^2+5\lambda-2$

A's eigenvalues are:

  • $\lambda_1 = \dfrac{-5+\sqrt{33}}{2}$

  • $\lambda_2 = \dfrac{-5-\sqrt{33}}{2}$

Now what do I do to solve the differential equation using A's eigenvector?

  • $\begingroup$ The solution to $x'=Ax,x(0)=x_0$ is $x(t)=e^{At}x_0$. If $A$ is diagonalizable with eigenvalue matrix $\Lambda$ and corresponding eigenvector matrix $V$ then $e^{At}=Ve^{\Lambda t}V^{-1}$ where $e^{\Lambda t}$ is computed by just taking exponentials of the diagonal entries. (Incidentally, this is the easiest way to compute $e^{At}$; I don't know what other way you had in mind.) So you need to find the eigenvectors (which you haven't done yet) and then plug into this formula. $\endgroup$ – Ian Aug 22 '16 at 15:03
  • $\begingroup$ I usually use an algorithm that follows Cayley's and Hamilton's theorem that is very straight forward $\endgroup$ – aribaldi Aug 22 '16 at 15:07
  • $\begingroup$ I believe I am supposed to solve it without using the exponential of the matrix is that possible? $\endgroup$ – aribaldi Aug 22 '16 at 15:08
  • $\begingroup$ In a sense it is. The matrix exponential solution boils down to: if $x_0=\sum_i c_i v_i$ then $e^{At}x_0=\sum_i c_i v_i e^{\lambda_i t}$, where $v_i$ are the eigenvectors, $\lambda_i$ are the corresponding eigenvalues, and $c_i$ are coefficients (which depend on $x_0$). In this way you can solve a particular IVP without writing down the matrix exponential in general. $\endgroup$ – Ian Aug 22 '16 at 15:09
  • $\begingroup$ Yes I believe that's how I should do it. I guess I would then just have to find the $c_i$ using the initial conditions. Thank You! $\endgroup$ – aribaldi Aug 22 '16 at 15:14

You have done most of the work already. $A's$ eigenvalues are:

$$\lambda_{1, 2} = \dfrac{-5\pm\sqrt{33}}{2}$$

We now have to find the eigenvectors and solve $[A - \lambda_i I]v_i = 0$. So for $\lambda_{1,2}$, we have:

$$[A - \lambda_{1,2} I]v_{1,2} = [A - \dfrac{1}{2} (-5 \pm \sqrt{33}) I]v_{1,2} = \begin{pmatrix} - \dfrac{1}{2} (-5 \pm \sqrt{33}) & 1 \\ 2 & -5 - \dfrac{1}{2}(-5 \pm \sqrt{33}) \end{pmatrix}v_{1, 2} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$

When we do the RREF of this matrix, we have:

$$\begin{pmatrix} 1 & \dfrac{1}{4} (-5 \mp\sqrt{33}) \\ 0 & 0 \end{pmatrix}v_{1, 2}= \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$

This gives us:

$$v_{1,2} = \begin{pmatrix} \dfrac{1}{4}(5 \pm \sqrt{33})\\1 \end{pmatrix}$$

We can now write:

$$X(t) = \begin{pmatrix} x(t)\\y(t) \end{pmatrix} = c_1 e^{\lambda_1 t} v_1 + c_2 e^{\lambda_2 t} v_2$$

You have two initial conditions to solve for $c_1$ and $c_2$ and should get:

$$X(t) = \begin{pmatrix} x(t)\\y(t) \end{pmatrix} =\begin{pmatrix} \dfrac{2}{\sqrt{33}}~e^{-1/2 (5+\sqrt{33}) t} (e^{\sqrt{33} t}-1) \\ \dfrac{1}{33} e^{-1/2 (5+\sqrt{33}) t} ((33-5 \sqrt{33}) e^{\sqrt{33} t}+33+5 \sqrt{33}) \end{pmatrix}$$

Update For $\lambda_1$, our RREF is:

$$\begin{pmatrix} 1 & \dfrac{1}{4} (-5 -\sqrt{33}) \\ 0 & 0 \end{pmatrix}v_1= \begin{pmatrix} 0 \\ 0 \end{pmatrix}$$

Let the vector $v_1 = \begin{pmatrix} a \\ b \end{pmatrix}$, so this gives us:

$$a + \dfrac{1}{4} (-5 -\sqrt{33})b = 0 \implies a = \dfrac{1}{4} (5 +\sqrt{33})b$$

We are free to choose whatever $b$ we'd like, so choose $b = 1$ and conclude.

This is how we arrived at:

$$v_1 = \begin{pmatrix} \dfrac{1}{4}(5 + \sqrt{33})\\1 \end{pmatrix}$$

I should also mention, note how this choice also satisfies the second equation.

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  • $\begingroup$ I don't understand how you got that form after performing the RREF $\endgroup$ – aribaldi Aug 23 '16 at 11:10
  • $\begingroup$ I added details for one of the eigenvalues after the RREF. $\endgroup$ – Moo Aug 23 '16 at 11:46

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