[proofplan] We bound the global error of Euler's method in three steps: compute the local truncation error via Taylor's theorem, derive a recursive inequality using the Lipschitz condition, and close the estimate with a discrete Gronwall lemma. [/proofplan] [step:Compute the local truncation error] By Taylor's theorem, $y(t_{n+1}) = y(t_n) + hy'(t_n) + \frac{h^2}{2}y''(\tau_n)$ for some $\tau_n \in (t_n, t_{n+1})$. Since $y'(t_n) = f(t_n, y(t_n))$, the local truncation error is $\eta_{n+1} = \frac{h^2}{2}y''(\tau_n)$. Set $c = \frac{1}{2}\max_{[0,T]}\|y''\|$. [/step] [step:Derive the error recursion using the Lipschitz condition] The global error $e_n = y_n - y(t_n)$ satisfies \begin{align*} e_{n+1} &= e_n + h\big(f(t_n, y_n) - f(t_n, y(t_n))\big) - \eta_{n+1}. \end{align*} Taking norms and using the Lipschitz condition: \begin{align*} \|e_{n+1}\| &\leq (1 + h\lambda)\|e_n\| + ch^2. \end{align*} [/step] [step:Apply the discrete Gronwall lemma] With $\|e_0\| = 0$ and the recursion $E_{n+1} \leq (1 + h\lambda)E_n + ch^2$, unrolling gives \begin{align*} \|e_n\| &\leq ch^2 \sum_{k=0}^{n-1}(1 + h\lambda)^k = ch^2 \cdot \frac{(1+h\lambda)^n - 1}{h\lambda}. \end{align*} Since $(1 + h\lambda)^n \leq e^{nh\lambda} \leq e^{\lambda T}$ for $nh \leq T$: \begin{align*} \|e_n\| &\leq ch \cdot \frac{e^{\lambda T} - 1}{\lambda}. \end{align*} [/step]