Solve the ODE $y''+4y'+3y = 4.5\sin(2t)$

Find the steady-state motion of the mass spring system modelled by the ODE $y''+4y'+3y = 4.5sin(2t)$

What I have tried:

Finding the general solution $$y^2+4y+3=0 \\ (y+1)(y+3) = 0 \\ \implies y_h = c_1e^{-x}+c_2e^{-3x}$$ To find $y_p$ we use $$y_p(t) = a\cos(\omega t) + b\sin(\omega t) \\ y'_p(t) = -a \omega \sin(\omega t) + b \omega \cos(\omega t) \\ y''_p(t) = -a \omega^2 \cos(\omega t) - b \omega^2 \sin(\omega t)$$

Plugging these into the equation: $$-\omega^2(a\cos(\omega t) + b\sin(\omega t))+4\omega(-a\sin(\omega t) + b\cos(\omega t)) + 3(a\cos(\omega t) + b\sin(\omega t))=4.5\sin(2t)$$ $\omega=2$ $$-2^2(a\cos(2 t) + b\sin(2t))+8(-a\sin(2t) + b\cos(2t)) + 3(a\cos(2t) + b\sin(2t))=4.5\sin(2t)$$

Simplifying I get $$-a\cos(2t)-3b\sin(2t)-8a\sin(2t)+8b\cos(2t)=4.5\sin(2t)$$

However, I'm not sure on how to proceed from here. I thought I'd need to use the calculations for $a$ and $b$ for the undetermined coefficients, however I $a = -\frac{1}{65}, b = \frac{4}{65}$ when plugged in I do not get the answer which is $y_p(t) = -36/65 \cos(2t)-9/130 \sin(2t)$

After your simplification you should end up with

$$-a\cos(2t)-b\sin(2t)-8a\sin(2t)+8b\cos(2t)=4.5\sin(2t)$$

careful because you had $-3b\sin\left(2t\right)$ and that was a little mistake. Now regrouping terms depending on $a$ and $b$ you get

$$ \left(-a+8b\right)\cos\left(2t\right)+\left(-8a-b\right)\sin\left(2t\right) = \frac{9}{2}\sin\left(2t\right) $$ You can then identify giving you $$ -a+8b = 0 \text{ and } -8a-b = \frac{9}{2} $$ I let you find the value of $a$ and $b$ which correspond to your reference answer