5. Use Algorithm 5.4 to approximate the solutions to the initial-value problems in Exercise 1
In Exercise 1
a. y = te3t − 2y, 0≤ t ≤ 1, y(0) = 0, with h=0.2; actual solution y(t) = 1/5 te3t - 1/25 e3t +1/25 e−2t .
b. y = 1 + (t − y)2, 2≤ t ≤ 3, y(2) = 1, with h = 0.2; actual solution y(t) = t + 1/1−t .
c. y = 1 + y/t, 1≤ t ≤ 2, y(1) = 2, with h = 0.2; actual solution y(t) = t ln t + 2t.
d. y = cos 2t + sin 3t, 0 ≤ t ≤ 1, y(0) = 1, with h = 0.2; actual solution y(t) = 1/2 sin 2t - 1/3 cos 3t + 4/3 .
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