Repeat Exercise 1 using the algorithm developed in Exercise 5.

In Exercise 1

a. u'₁ = 3u₁ + 2u₂ − (2t² + 1)e^2t , u₁(0) = 1; u'₂ = 4u₁ + u₂ + (t² + 2t − 4)e^2t , u₂(0) = 1; 0 ≤ t ≤ 1; h = 0.2; actual solutions u₁(t) = 1/3 e^5t – 1/3 e^−t + e^2t and u₂(t) = 1/3 e^5t + 2/3 e^−t + t²e^2t .

b. u'₁ = −4u₁ − 2u₂ + cos t + 4 sin t, u₁(0) = 0; u'₂ = 3u₁ + u₂ − 3 sin t, u₂(0) = −1; 0 ≤ t ≤ 2; h = 0.1; actual solutions u₁(t) = 2e^−t − 2e^−2t + sin t and u₂(t) = −3e^−t + 2e^−2t .

c. u'₁ = u₂, u₁(0) = 1; u'₂ = −u₁ − 2e_t + 1, u₂(0) = 0; u'₃ = −u₁ − e^t + 1, u₃(0) = 1; 0 ≤ t ≤ 2; h = 0.5; actual solutions u₁(t) = cos t + sin t − e^t + 1, u₂(t) = −sin t + cos t − e^t , and u₃(t) = −sin t + cos t.

d. u'₁ = u₂ − u₃ + t, u₁(0) = 1; u'₂ = 3t², u₂(0) = 1; u'₃ = u₂ + e^−t , u₃(0) = −1; 0 ≤ t ≤ 1; h = 0.1; actual solutions u₁(t) = −0.05t⁵ + 0.25t⁴ + t + 2 − e^−t , u₂(t) = t³ + 1, and u₃(t) = 0.25t4 + t − e^−t .