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We consider the following time stepping methods for an initial value problem y(t) = f(t, y(t)) with time step size h = ti+1 ti: Method A: Yi+1 = Yi + hf (tit wa - ) h Yi + Yi+1 2' 2 Method B: Yi+1 = Yi + hf ( ti + 2 h Yi + 5 f(ti, yi) (a) Using the problem y(t) = 10 cos(t) 2y(t) with y(0) 1 and 0