Answer:
Explanation:
xy = 2y + xy = 0
Hence, 2y + xy = 0 ---------(1)
Differentiating equation (1) n times by Leibnitz theorem, gives:
2y(n) + xy(n) + ny(n - 1) = 0
Let x = 0: 2y(n) + ny(n - 1) = 0
2y(n) = -ny(n - 1)
∴ y(n) = -ny(n - 1)/2 for n ≥ 1
For n = 1: y = 0
For n = 2: y(1) = -y
For n = 3: -3y(2)/2
For n = 4: -2y(3)