# I'm asked to decide if I should solve the system dot y=(−600 400 400 −600)y, t in [t_0,t_e], y(t_0)=y_0 with either the explicit Euler method or the implicit Euler method.

I'm asked to decide if I should solve the system
$\stackrel{˙}{y}=\left(\begin{array}{cc}-600& 400\\ 400& -600\end{array}\right)y,\phantom{\rule{1em}{0ex}}t\in \left[{t}_{0},{t}_{e}\right],\phantom{\rule{1em}{0ex}}y\left({t}_{0}\right)={y}_{0}$
with either the explicit Euler method or the implicit Euler method.

Using the explicit Euler method I would get the updating scheme
${y}_{n+1}=\left(\begin{array}{cc}1-600h& 400h\\ 400h& 1-600h\end{array}\right){y}_{n}$
where the eigenvalues of the driving matrix is
${\lambda }_{1}=401-600h,$
${\lambda }_{2}=-399-600h.$
For the solution to be stable these need to be less than one which gives the conditions
$h\ge \frac{4}{6}=\frac{2}{3},$
$h\ge -\frac{4}{6}=-\frac{2}{3}.$
The last condition doesn't say anything but the first condition seems restrictive since I can't choose $h$ as small as possible.

If I instead were to use the implicit Euler method I would get the updating scheme
${y}_{n+1}={\left(\begin{array}{cc}1-600h& 400h\\ 400h& 1-600h\end{array}\right)}^{-1}{y}_{n}.$
Now I can't solve for the eigenvalues of this system but I've heard the implicit Euler is unconditionally stable so it shouldn't matter.

So is the answer that I should choose implicit Euler because it is unconditionally stable or am I missing something? The order of consistency of both is 1 so that should not matter.
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Abraham Norris
Your computation of eigenvalues is wrong. Of the system matrix in ${y}^{\prime }=Ay$ there is one eigenvalue −200 with eigenvector $\left(\begin{array}{c}1\\ 1\end{array}\right)$ and one eigenvalue −1000 with eigenvector $\left(\begin{array}{c}1\\ -1\end{array}\right)$. These translate into eigenvalues $1-200h$ and $1-1000h$ of the "driving matrix" for the Euler method, requiring $h<0.002$ for stability.
For the implicit method there are no step size restrictions to get stability in the method, to get into the range where the error behaves like order 1 one still will need $500h\ll 1$.