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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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closed as unclear what you're asking by Did, Travis, Ali Caglayan, Newb, Mark Fantini Jan 18 '15 at 15:09

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

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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