Problem Statement:
Given the initial-value problem $\displaystyle \frac{\ 2*y}{t} + t^2e^t, 1\leq t \leq 2, y(1) = 0,$ with exact solution $\displaystyle y(t) = t^2(e^t-e):$ Compute the value of h necessary for $\displaystyle \mid y(t_i) - w_i \mid \leq 0.1$ using this equation $\displaystyle \mid y(t_i) - w_i \mid \leq \frac{\ hM}{2L}(e^{L(t_i-a)}-1)$.

Solution so far.

I think L= 1 because $\displaystyle y' = \frac{\ 2*y}{t} + t^2e^t$ is differentiable for every y, but I am not completely sure. Also I am unsure on how to find M. The textbook gives this relationship: $\displaystyle \mid y''(t)\mid \leq M$. I found that the second derivative to be $\displaystyle y''(t) = \frac{\ 2*y'-2y}{t^2}+2te^t+t^2*e^t$. I plugged in the highest t, 2, and the corresponding y, 18.6831. The y''(t) = 98.0102.
a = 1
h is the unknown.