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For $f(x,y)=y$ and $\beta =1/2$, the explicit midpoint method multiplies the current value by $R=1+h+h^2/2$. Starting from one, induction givesLet $A=e^h$ and $B=R$. The Taylor theorem with Lagrange remainder for the exponential function givesThe difference of powers identity gives $A^n-B^n=(A-B){\sum }_{k=0}^{n-1}{A}^{n-1-k}{B}^k$. Every summand lies between one and $A^{n-1}$. Since $nh=x_n$, the global error therefore satisfiesThe same inequalities hold at $n=0$, where the error vanishes. At the fixed endpoint $x_N=1$, the lower bound $h^2/6$ prevents any higher order of a numerical method, while the upper bound is $e{h}^2/6$. Thus the convergence order is exactly two.