Use the Extrapolation Algorithm with tolerance TOL = 10−4, hmax = 0.25, and hmin = 0.05 to approximate the solutions to the following initial-value problems. Compare the results to the actual values.
a. y' = te3t − 2y, 0≤ t ≤ 1, y(0) = 0; actual solution y(t) = 1/5 te3t - 1/25 e3t + 1/25 e−2t .
b. y' = 1 + (t − y)2, 2≤ t ≤ 3, y(2) = 1; actual solution y(t) = t + 1/(1 − t).
c. y' = 1 + y/t, 1≤ t ≤ 2, y(1) = 2; actual solution y(t) = t ln t + 2t.
d. y' = cos 2t + sin 3t, 0≤ t ≤ 1, y(0) = 1; actual solution y(t) = 1/2 sin 2t - 1/3 cos 3t + 4/3 .