How to solve $y''(t) + 4 y (t) = 2t \cos (2t)$ on time in the exam? I'm solving this ODE, which is from last year final exam

Solve in $\mathbb R$ the ODE $$y''(t) + 4 y (t) = 2t \cos (2t)$$

This ODE is not hard to solve, but it takes me a lot of time to compute the derivative $y',y''$ and simplifies them. This process is unfortunately likely to cause mistake.
The last year exam contains a total of $4$ questions, and has a duration of $1$ hour and $30$ minutes. Solving this ODE is just one of $4$ questions.

I would like to ask for a faster way to solve this ODE. Maybe it has a special characteristics that I'm unable to recognize.


My attempt:
The solution is of the form $$\begin{aligned} y (t) &= \big ( a \cos (2t) + b \sin (2t) \big ) + t(ct+d) \cos (2t) + t(gt+h) \sin (2t)\\
&= a \cos (2t) + b \sin (2t) + dt \cos (2t) +ht \sin (2t) + ct^2 \cos (2t) + gt^2 \sin (2t) \end{aligned}$$
where $a,b,c,d,g,h \in \mathbb R$. It follows that $$\begin{aligned}
y' (t) &= (2b+d) \cos (2t) + (h-2a) \sin (2t)\\
       & \qquad + 2(h+c) t \cos (2t) +2(g-d) t \sin (2t)\\
       & \qquad + 2g t^2 \cos (2t)  -2c t^2 \sin (2t)
\end{aligned}$$ and $$\begin{aligned}
y'' (t) &= 2(2h-2a+c) \cos (2t) -2 (2b+2d-g) \sin (2t)\\
       & \qquad + 4(2g-d) t \cos (2t) -4(h+2c) t \sin (2t)\\
       & \qquad -4c t^2 \cos (2t)  -4g t^2 \sin (2t)
\end{aligned}$$
Hence $$\begin{cases}
2(2h-2a+c) + 4a &=0 \\
-2 (2b+2d-g) +4b &=0 \\
4(2g-d) + 4d &=0 \\
-4(h+2c) + 4h  &=0 \\
-4c +4c &=0 \\
-4g+4g  &=0 \\
\end{cases} \iff
\begin{cases}
c &=0 \\
g &=1/4 \\
h &=0 \\
d  &=1/8 \\
\end{cases}$$
In conclusion, the solution is $$y(t) = a \cos (2t) + b \sin (2t) + \frac{1}{8} t \cos (2t)   + \frac{1}{4} t^2 \sin (2t)$$

You can reduce the effort to one half by considering $y$ as the real part of the complex-valued solution to 
$$
z''+4z=2te^{2it}
$$
Then one particular solution is of the form $z_p(t)=t(At+B)e^{2it}$ by the method of undetermined coefficients, and the root structure that you correctly found. Now inserting this gives simply a $2\times 2$ linear system to solve,
\begin{align}
z_p'(t)&=(2At+B+2i(At^2+Bt))e^{2it}\\
z_p''(t)&=(4iAt+2(A+iB)-4At^2+4i(A+iB)t+2iB)e^{2it}\\
\hline
2t&=[-4At^2+4i(2A+B)t+2(A+2iB)] + 4[At^2+Bt]\\
&=8iAt+2(A+2iB)\\
\implies A&=-\frac{i}4,~~B=\frac{i}2A=\frac18
\end{align}
and thus
$$
y_p(t)=Re(z_p(t))=\frac{t^2}4\sin(2t)+\frac{t}8\cos(2t)
$$
