# Use all the Adams-Bashforth methods to approximate the solutions to the following initial-value problems. In each case use exact starting values, and compare the results to the actual values. 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 e2t . b. y' = 1

Use all the Adams-Bashforth methods to approximate the solutions to the following initial-value problems. In each case use exact starting values, and compare the results to the actual values.

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 .

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 .

## This problem has been solved!

Do you need an answer to a question different from the above? Ask your question!

**Related Book For**