$y_p$ of $y''+4y=3\sin2t$? I was trying to solve the nonhomogeneous equation 
$$
y''+4y=3\sin2t, \\
y(0)=0, \\
y'(0)=-1
$$
by solving the homogeneous equation I got $y_h= C_1\sin(2t) + C_2\cos2t$
and then I set $y_p= A\sin2t + B\cos2t$ and found the double derivative which I found to be $-4A\sin(2t) - 4B\cos(2t)$ so I substituted the second derivative for $y''$ and $y_p$ for $4y$ and then I got $0A + 0B = 3\sin(2t)$ this cant be right? so what I was wondering is how do I find the correct $y_p$ and also is it somehow to decide early that my $y_p$ wont work? I mean maybe my $y_p$ is correct, I might have done a critical mistake with the derivatives or something but I doubt it, any tips/solutions? thanks  
 A: Notice that the form you took for $y_p$ is already accounted for in $y_h$ (in fact the two forms are identical!). When this happens, you need to try multiplying by a power $t$:
$$y_p=t(A\cos(2t)+B\sin(2t))$$
which, for a suitable power of $t$, will result in a $y_p$ that is linearly independent from $y_h$.
In general, for the method of undetermined coefficients, when $y_p$ is already accounted for in $y_h$, you use $\tilde{y}_p=t^ky_p$ where $k$ is the smallest power for which $\tilde{y}_p$ is not accounted for in $y_h$.
The idea is the general solution of the given problem is of the form $y=y_h+y_p$, but if $y_p=Ky_h$ (i.e. $y_p$ and $y_h$ are linearly dependent), then $y_p$ adds nothing to the general solution $y$ that wasn't already "there" in $y_h$.
A precise discussion of this (and why it works) is found in every elementary ODE text. This "multiplying by a power of $t$" on the "standard first guess" $y_p$ in scenarios like this is the reason behind the physical phenomenon of resonance in, e.g., undamped, spring-mass systems.
