structural:summarization, structural:reformatting
Example

If we have the differential equation y ′ = 1 + (t − y) 2 {\displaystyle y'=1+(t-y)^{2}}, and the exact solution y = t + 1 t − 1 {\displaystyle y=t+{\frac {1}{t-1}}}, and we want to find M {\displaystyle M} and L {\displaystyle L} for when 2 ≤ t ≤ 3 {\displaystyle 2\leq t\leq 3}. L = max (| d d y |) = max 2 ≤ t ≤ 3 (| d d y |) = max 2 ≤ t ≤ 3 (| 2 (t − y) |) = max 2 ≤ t ≤ 3 (| 2 (t −) |) = max 2 ≤ t ≤ 3 (| − 2 t − 1 |) = 2 {\displaystyle L={\text{max}}{\bigl (}|{\frac {d}{dy}}{\bigl }|{\bigr)}=\max _{2\leq t\leq 3}{\bigl (}|{\frac {d}{dy}}{\bigl }|{\bigr)}=\max _{2\leq t\leq 3}{\bigl (}|2(t-y)|{\bigr)}=\max _{2\leq t\leq 3}{\bigl (}|2(t-)|{\bigr)}=\max _{2\leq t\leq 3}{\bigl (}|-{\frac {2}{t-1}}|{\bigr)}=2} M = max (| d 2 d t 2 |) = max 2 ≤ t ≤ 3 (| d 2 d t 2 |) = max 2 ≤ t ≤ 3 (| 2 (− t + 1) 3 |) = 2 {\displaystyle M={\text{max}}{\bigl (}|{\frac {d^{2}}{dt^{2}}}{\bigl }|{\bigr)}=\max _{2\leq t\leq 3}\left(|{\frac {d^{2}}{dt^{2}}}{\bigl }|\right)=\max _{2\leq t\leq 3}\left(|{\frac {2}{(-t+1)^{3}}}|\right)=2} Thus we can find the error bound at t=2.5 and h=0.5:

error bound = h M 2 L (e L (t i − t 0) − 1) = 0.5 ⋅ 2 2 ⋅ 2 (e 2 (2.5 − 2) − 1) = 0.42957 {\displaystyle {\text{error bound}}={\frac {hM}{2L}}\left(e^{L(t_{i}-t_{0})}-1\right)={\frac {0.5\cdot 2}{2\cdot 2}}\left(e^{2(2.5-2)}-1\right)=0.42957} Notice that t is equal to 2 because it is the lower bound for t in 2 ≤ t ≤ 3 {\displaystyle 2\leq t\leq 3}.