Solve DE using variable separable

Question

(1 + cos 2y) dt - 2t² dy =0, y(1) = π/4

a. y = Arctan (2 - 1/t)

b. y = Arctan (t - 1)

c. y = Arctan (1/t - 2)

d. y = Arctan (2 - t)

Answer 1

dy/(1 + cos2y) = dt/(2t²); ∫dy/cos²y= ∫dt/t^2; tany = -1/t + C; tanπ/4 = - 1 + C; C = 2; tany = 2 - 1/t;

y = Arctan(2 - 1/t);

Answer is a.