The actual solutions to the initial-value problems in Exercise 6 are given here. Compute the actual error in the approximations of Exercise 6.
a. y(t) = (2t + 1)/(t2 + 1)
b. y(t) = −1/(ln(t + 1))
c. y(t) = 2t(1 − 2t)
d. y(t) = √(4 − 3e−t2)