Diffeq question: method of undetermined coefficients 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.

Your problem is $(D^2+4)[y] = 3 \sin (2t)$. Operate by $D^2+4$ once again to find $(D^2+4)^2[y] = (D^2+4)(3\sin(2t)) =0$. Then,
$$ y = c_1 \cos( 2t)+c_2\sin( 2t) + c_3t\cos (2t) + c_4t \sin (2t) $$
is the general solution. Identify the first two terms as the homogeneous solution to the given differential equation. Hence form:
$$ y_p = t(A \cos (2t)+ \sin (2t)) $$
via the $c_3$ and $c_4$ terms. Next, 
$$ (D^2+4)[y_p] = 3 \sin (2t)$$
will yield equations to fix $A$, $B$. Finally, apply the initial conditions to fix $c_1,c_2$.
