Repeat Exercise 2 using the Adams fourth-order predictor-corrector method.

In Exercise 2

a. y' = −5y + 6e^t, 0≤ t ≤ 1, y(0) = 2, with h = 0.1; actual solution y(t) = e^−5t + e^t .

b. y' = −10y+10t+1, 0 ≤ t ≤ 1, y(0) = e, with h = 0.1; actual solution y(t) = e^−10t+1+t.

c. y' = −15(y − t⁻³) − 3/t⁴, 1 ≤ t ≤ 3, y(1) = 0, with h = 0.25; actual solution y(t) = −e^−15t + t⁻³.

d. y' = −20y + 20 cost − sint, 0 ≤ t ≤ 2, y(0) =0, with h = 0.25; actual solution y(t) = −e^−20t + cos t.