Repeat Exercise 2 using the algorithm developed in Exercise 5.

In Exercise 2

a. u'₁ = u₁ − u₂ + 2, u₁(0) = −1; u'₂ = −u₁ + u₂ + 4t, u₂(0) = 0; 0 ≤ t ≤ 1; h = 0.1; actual solutions u₁(t) = −1/2 e^2t + t² + 2t – ½ and u₂(t) = 1/2 e^2t + t² – 1/2.

b. u'₁ = 1/9 u₁ – 2/3 u₂ – 1/9 t² + 2/3 , u₁(0) = −3; u'₂ = u₂ + 3t − 4, u₂(0) = 5; 0 ≤ t ≤ 2; h = 0.2; actual solutions u₁(t) = −3et + t2 and u₂(t) = 4e^t − 3t + 1.

c. u'₁ = u₁ + 2u₂ − 2u₃ + e^−t , u₁(0) = 3; u'₂ = u₂ + u₃ − 2e−t , u₂(0) = −1; u'₃ = u₁ + 2u₂ + e^−t , u₃(0) = 1; 0 ≤ t ≤ 1; h = 0.1; actual solutions u₁(t) = −3e^−t −3 sin t +6 cos t, u₂(t) = 3/2 e^−t + 3/10 sin t – 21/10 cos t – 2/5 e^2t , and u₃(t) = −e^−t + 12/5 cos t + 9/5 sin t – 2/5 e^2t .

d. u'₁ = 3u₁ + 2u₂ − u₃ − 1 − 3t − 2 sin t, u₁(0) = 5; u'₂ = u₁ − 2u₂ + 3u₃ + 6 − t + 2 sin t + cos t, u₂(0) = −9; u'₃ = 2u₁ + 4u₃ + 8 − 2t, u₃(0) = −5; 0 ≤ t ≤ 2; h = 0.2; actual solutions u₁(t) = 2e^3t + 3e^−2t + 1, u₂(t) = −8e^−2t + e^4t − 2e^3t + sin t, and u₃(t) = 2e^4t − 4e^3t − e^−2t − 2.