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.

  • $\begingroup$ Writint out a Taylor polynomial would ideed be a good idea, I think. $\endgroup$
    – Matti P.
    May 22, 2019 at 7:14
  • 1
    $\begingroup$ 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. $\endgroup$ May 22, 2019 at 7:24
  • $\begingroup$ 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. $\endgroup$
    – Dovendyr
    May 22, 2019 at 9:25
  • $\begingroup$ Simple order 2 methods are absolutely suited for a beginner course. Besides, this is more an exercise in Taylor polynomials than in numerical integration. $\endgroup$ May 22, 2019 at 9:31
  • $\begingroup$ I still don't understand, can you please show what you mean and which parts I am supposed to develop with Taylor? $\endgroup$
    – Dovendyr
    May 26, 2019 at 7:24

1 Answer 1


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 $α=0$.

You will need to confirm that this holds true for any differential equation, not just this (quite standard) test case.


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