I have a question here from Differential Equations. We are learning how to solve second order nonhomogenous equations using the method of undetermined coefficients. The equation at hand is $$y''+4y=3\sin(2t)$$ I understand that the solution to the corresponding homogenous equation is $$y_c(t)=c_1\cos(2t)+c_2\sin(2t)$$ And I understand that to solve the equation using this method, I have to find an equation with "undetermined coefficients" that I then add to $y_c(t)$. But what form does this part of the equation take?
p.s. The answer in the back of the book is $y=2\cos2t-(1/8)\sin2t-(3/4)t\cos2t$. The initial conditions were $y(0)=2, y'(0)=-1$, but I don't need help with initial conditions.