Solve initial value problem

I have an exercise that I do not understand. We have to solve an initial value problem:

$$\begin{array}{ccl} y'(t) &=& f(t,y(t)) \\ y(a) &=& y_0 \end{array}$$

We have to derive an effective method to calculate $$\alpha$$, $$\beta$$ and $$\gamma$$ with the highest possible accuracy. The method has the form below: $$y_{k+1} = \alpha y_k + \beta y_{k-1} + \gamma h f(t_k, y_k)$$ We can even assume that $$y_1 = y(a+h)$$ and that the interval is equidistant.

I can guess that we have to derive a Euler forward, but I cannot solve it? Besides, they ask the order of accuracy of the found method. Shall I write all the $$y(a+h)$$ as a Taylor polynomial? I would greatly appreciate some guidance.

• Writint out a Taylor polynomial would ideed be a good idea, I think. – Matti P. May 22 at 7:14
• No, it will not be Euler forward, the resulting method is a second order two-step midpoint or Nyström method, like in this question, further explored here. – LutzL May 22 at 7:24
• I don't think it's a Nyström or modified Euler, it's a beginner course and we did not do those methods. I tried to make a Taylor polynomial but nothing disappears. – Dovendyr May 22 at 9:25
• Simple order 2 methods are absolutely suited for a beginner course. Besides, this is more an exercise in Taylor polynomials than in numerical integration. – LutzL May 22 at 9:31
• I still don't understand, can you please show what you mean and which parts I am supposed to develop with Taylor? – Dovendyr May 26 at 7:24

For the test problem $$f(t,y)=y$$ you get the exact solution $$y=Ce^x$$ and with the constant chosen so that $$y_k=1$$ the equation reduces to $$e^h=α+βe^{-h}+γh$$ which has to be correct up to some power of $$h$$.
Inserting the power series of the exponential and comparing coefficients of equal degree gives \begin{align} γh&=(1+h+\tfrac12h^2+\tfrac16h^3+...)-α-β(1-h+\tfrac12h^2-\tfrac16h^3\pm...) \\[1em]\hline 0&=1-α-β\\ γ&=1+β\\ 0&=\tfrac12(1-β)\\ 0&=\tfrac16(1+β) \end{align} which have to be satisfied from top down as far as possible. Obviously, the last equation contradicts the next-to-last one. Thus $$β=1$$, $$γ=2$$ and $$α=0$$.