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Below is a problem I recited from an exam I took. I wasn't able to solve it on time. Could someone show me how to simplify it?

If $$f(t,y(t)) = y'(t)$$ $$y'(t) = -\lambda y(t)$$ where $\lambda$ is a positive constant.

$$y_{n+1} = y_{n} + \frac{h}{2}(f(t_n,y_n) + f(\frac{2}{3}h + t_{n},y_{n} +f(t_{n-1},y_{n-1})))$$

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I'll assume you know the theory, so I'll keep it fairly short (ask if you need more details). The method applied to the linear differential equation $y' = -\lambda u$ simplifies to $y_{n+1} = (1-k)y_n + \frac12k^2y_{n-1}$ with $k=h\lambda$; here I assume that you missed out a factor $h$ in front of the final $f$ in your equation for the numerical method. This is a linear recurrence relation with characterictic polynomial $a^2 - (1-k)a - \frac12k^2$, and the method is stable if the roots of this polynomial are within the unit disc. The discriminant of the quadratic polynomial is $1-2k+3k^2$, which is never negative as $1-2k+3k^2 \ge 1-2k+k^2 = (1-k)^2 \ge 0$, so we have two real roots. Thus, the roots can only cross the unit circle by going through $\pm1$. Substitute $a=1$ in the characteristic polynomial yields $k=2$, and the case $a=-1$ yields $k=-1+\sqrt{5}$, so the method is stable for $k \le -1+\sqrt{5}$.

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If you have $y'(t) = -\lambda y(t)$, then you have a linear ordinary differential equation. You have tagged this question "numerical analysis"; as a consequence, I will address this from a numerical standpoint.

Consider an explicit one-step method. One-step methods can be organized by "order"; a first-order one step method uses a single approximation of the solution in the forward direction; a second-order method essentially divides the interval into two segments, and computes the forward step from there. The most common one-step method is the 4th order Runge-Kutta method, which has a fairly straightforward form.

For all of these methods, there is a region of absolute stability. That is, if you have a linear ODE, if the eigenvalues times the step size $h$ fall within this region, then you are guaranteed to converge on a solution.

You have not specified which method you are asked to analyze. However, it is clear to see that the eigenvalue of your problem is simply $\lambda$; therefore, $\lambda$ must fall within the region of absolute stability for your method.

All explicit one-step methods completely enclose the region of absolute stability for lower-order methods. For Euler's explicit method, the first-order technique, the region of absolute stability is a disk in the complex plane with radius 1 centered at $z = -1$.

Therefore, $h\lambda$ must be within this disk.

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