I have:
$$y' = -22*y+3*sin(3*x)$$ $$0 \le x \le 3 $$ $$y(0) = 4$$
as my initial value problem.
The question is:
What is the largest step size h for which the Euler method is stable, when applied to the initial value problem above?
I believe the region of stability for Euler is something like:
$$|1+h*\lambda| < 1$$
Don't really get what $\lambda$ is meant to be?
I also have a follow-up question:
When applying the same theoretical analysis to the midpoint method, what is the largest step size h for which the method is stable?
PS. $\lambda$ is a constant in the complex numbers and I believe $Re \lambda < 0$.
We have
$$ \frac{y_{k+1}-y_k}{h} + 22 y_k = 3\sin(3 k h) $$
The method stability is related to the stability for the recurrence
$$ \frac{y_{k+1}-y_k}{h} + 22 y_k = 0 $$
Solving the homogeneous linear recurrence equation we have
$$ y_k = C_0(1-22h)^{k-1} $$
which is stable for $|1-22h| < 1$ or $-1 < 1-22h < 1$ or
$$ 0 < h <\frac{1}{11} $$
NOTE
Solving the recurrence
$$ \frac{y_{k+1}-y_k}{h} + 22 y_k = 3\sin(3 k h) $$
with $C_0 = 0$ we have
$$ y_k = \frac{3 h \left(\sin (3 h) \left((1-22 h)^k-\cos (3 h k)\right)+(22 h+\cos (3 h)-1) \sin (3 h k)\right)}{44 h (11 h-1)+(44 h-2) \cos (3 h)+2} $$
Follows a plot showing the result for $h = \frac{1}{11}-0.0002$ (red) and $h = \frac{1}{11}-0.02$ (blue). Both solutions are stable but the blue shows better behavior being more resistant to limited precision and round-off errors.