(Solved): {y=m[y2e2t(t2+1)2]y+2tety(0)=1, whose true solution is y(t)=et(t2+1) for al ...

$\{\begin{array}{l}{y}^{\prime }=-m\left[y^2-e^{-2t}{\left(t^2+1\right)}^2\right]-y+2t{e}^{-t}\\ y(0)=1\end{array},$ whose true solution is $y(t)=e^{-t}\left(t^2+1\right)$ for all values of $m$. For [Q1], let $m=1$ and step size $h=0.2,t_k=kh(0\le k\le N)$ (a) Evaluate $y_1\approx y\left(t_1\right)$ obtained by the forward Euler's method. (b) Evaluate $y_1\approx y\left(t_1\right)$ obtained by the implicit trapezoid method (note that CrankNicolson in our textbook refers to the implicit midpoint method).