 Solution

For $h\lambda /2<1$, the implicit step map $z\mapsto y_n+(h/2)[f(t_n,y_n)+f(t_{n+1},z)]$ is a contraction, so the Banach fixed-point theorem supplies a unique next value. Smoothness of the exact solution gives the one-step defectuniformly on $[0,T]$, by the trapezoidal rule error estimate or a direct Taylor expansion. Let $e_n=y_n-y(t_n)$. Subtracting exact and numerical steps and applying the Lipschitz condition givesPut $r_h=(1+h\lambda /2)/(1-h\lambda /2)$. With $e_0=0$, iteration yields $|e_n|\le [C{h}^3/(1-h\lambda /2)](r_h^n-1)/(r_h-1)$. Since $r_h-1=h\lambda /(1-h\lambda /2)$, this is $C{h}^2(r_h^n-1)/\lambda$. For sufficiently small $h$, $\log r_h\le 2\lambda h$, so $r_h^n\le e^{2\lambda T}$ whenever $nh\le T$. ThereforeThis proves convergence with global order two directly from the defect bound and stability recurrence.