Repeat Exercise 1 using the Midpoint method.

In Exercise 1

a. y' = te^3t − 2y, 0≤ t ≤ 1, y(0) = 0, with h=0.5; actual solution y(t)=1/5 te^3t – 1/25 e^3t + 1/25 e^−2t .

b. y' = 1 + (t − y)2, 2≤ t ≤ 3, y(2) = 1, with h = 0.5; actual solution y(t) = t + 1/(1−t) .

c. y' = 1 + y/t, 1≤ t ≤ 2, y(1) = 2, with h = 0.25; actual solution y(t) = t ln t + 2t.

d. y' = cos 2t + sin 3t, 0 ≤ t ≤ 1, y(0) = 1, with h = 0.25; actual solution y(t) = 1/2 sin 2t – 1/3 cos 3t + 4/3 .