(4 pts) 16) Briefly explain what is wrong with the following attempt to solve the linear differential equation y' + 3y = sin x For this linear DE, P(x) = 3 and Q(x) = sin x The integrating factor is I(x) = eJ P(@)dx _ J 3dx - 3x Multiplying both sides by e3 then gives esty' + 3ety = sin x The left side is a product rule with F = est and S = y so integrating both sides gives esty = - cosa + C Finally, dividing by e3 gives y = - COS X e3x + Ce-3x(3 pts) 15) Is the function y = 2 cosa a solution to the differential equation y" + 2y' - 3y = -2 sinx - 6 sin x? (be sure to justify your answer!)