Consider the IVP

$\begin{array}{r}{y}^{\u2033}=-y\end{array}$

for $t\ge 0$, and $y(0)=1$, ${y}^{\prime}(0)=2$.

I have rewritten this differential equation as a system of first-order ODE's such that

$\begin{array}{r}{u}^{\prime}=v\\ {v}^{\prime}=-u\end{array}$

with $u(0)=1,v(0)=2$.

The solution is $y=2\mathrm{sin}(t)+\mathrm{cos}(t)$, ${y}^{\prime}=2\mathrm{cos}(t)-\mathrm{sin}(t)$

I am asked to perform one step of Euler's method with $h=0.5$, and determine if Euler's method is stable for this problem.

For the first part, I find that one step of Euler's method yields

$\begin{array}{r}{y}_{1}={y}_{0}+hf({t}_{0},{y}_{0})=1+(0.5)(2\mathrm{cos}(0)-\mathrm{sin}(0))=2.\end{array}$

But how do I determine if Euler's method is stable? I know that for the equation ${y}^{\prime}=\lambda y$, Euler's method is stable for $|1+h\lambda |\le 1$, but since this problem is in a different form I'm not sure what I need to do here.

$\begin{array}{r}{y}^{\u2033}=-y\end{array}$

for $t\ge 0$, and $y(0)=1$, ${y}^{\prime}(0)=2$.

I have rewritten this differential equation as a system of first-order ODE's such that

$\begin{array}{r}{u}^{\prime}=v\\ {v}^{\prime}=-u\end{array}$

with $u(0)=1,v(0)=2$.

The solution is $y=2\mathrm{sin}(t)+\mathrm{cos}(t)$, ${y}^{\prime}=2\mathrm{cos}(t)-\mathrm{sin}(t)$

I am asked to perform one step of Euler's method with $h=0.5$, and determine if Euler's method is stable for this problem.

For the first part, I find that one step of Euler's method yields

$\begin{array}{r}{y}_{1}={y}_{0}+hf({t}_{0},{y}_{0})=1+(0.5)(2\mathrm{cos}(0)-\mathrm{sin}(0))=2.\end{array}$

But how do I determine if Euler's method is stable? I know that for the equation ${y}^{\prime}=\lambda y$, Euler's method is stable for $|1+h\lambda |\le 1$, but since this problem is in a different form I'm not sure what I need to do here.