Use the Extrapolation Algorithm with tolerance TOL = 10−6, hmax = 0.5, and hmin = 0.05 to approximate the solutions to the following initial-value problems. Compare the results to the actual values.
a. y' = y/t − (y/t)2, 1≤ t ≤ 4, y(1) = 1; actual solution y(t) = t/(1 + ln t).
b. y' = 1 + y/t + (y/t)2, 1≤ t ≤ 3, y(1) = 0; actual solution y(t) = t tan(ln t).
c. y' = −(y + 1)(y + 3), 0≤ t ≤ 3, y(0) = −2; actual solution y(t) = −3 + 2(1 + e−2t)−1.
d. y' = (t + 2t3)y3 − ty, 0≤ t ≤ 2, y(0) = 1/3 ; actual solution y(t) = (3 + 2t2 + 6(et)2)−1/2.