The Laplace transform 5eᵗ - 2 / 6 exists, but without finding it solve the initial-value problem.
a. y(0) = 5, y'(0) = -2/6
b. y(0) = -2/6, y'(0) = 5
c. y(0) = 5, y'(0) = 2/6
d. y(0) = 2/6, y'(0) = 5