Repeat Exercise 2 using the algorithm developed in Exercise 5. In Exercise 2 a. u' 1 =
Question:
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 e2t + t2 + 2t – ½ and u2(t) = 1/2 e2t + 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) = 4et − 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 e2t , and u3(t) = −e−t + 12/5 cos t + 9/5 sin t – 2/5 e2t .
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) = 2e3t + 3e−2t + 1, u2(t) = −8e−2t + e4t − 2e3t + sin t, and u3(t) = 2e4t − 4e3t − e−2t − 2.
Step by Step Answer: