I have the following differential equation: y+y=cos⁡(t)cos⁡(2t) Maybe something can be done to cos⁡(t)cos⁡(2t) to...

I don't see an elegant solution , but it can be done by brute force using the method of variation of parameters.
The homogeneous differential equation ${\displaystyle y{}^{″}\left(t\right)+y\left(t\right)=0}$ has the two fundamental solutions ${\displaystyle y_1\left(t\right)=\sin \left(t\right)}$ and ${\displaystyle y_2\left(t\right)=\cos \left(t\right)}$
We must solve
$$\left(\begin{array}{c}0\\ \cos (t)\cos (2t)\end{array}\right)=\left(\begin{array}{cc}{y}_1& y_2\\ y_1^{\prime }& y_2^{\prime }\end{array}\right)\left(\begin{array}{c}{u}_1\\ u_2\end{array}\right)=\left(\begin{array}{cc}\sin t& \cos t\\ -\cos t& \sin t\end{array}\right)\left(\begin{array}{c}{u}_1\\ u_2\end{array}\right)$$
for ${\displaystyle u_1}$ and ${\displaystyle u_2}$ and find ${\displaystyle U_1}$ and ${\displaystyle U_2}$ such that ${\displaystyle U_1^{\prime }=u_1}$ and ${\displaystyle U_2^{\prime }=u_2}$
A particular solution will then be given by
${\displaystyle y_0\left(t\right)=U_1{y}_1+U_2{y}_2}$ and ${\displaystyle y=y_0+c_1{y}_1+c_2{y}_2}$
will be the general solution.
Since the matrix $A(t)=\left(\begin{array}{cc}\sin t& \cos t\\ -\cos t& \sin t\end{array}\right)$ is orthogonal, its inverse is its transpose, so
$$\left(\begin{array}{c}{u}_1\\ u_2\end{array}\right)=\left(\begin{array}{cc}\sin t& -\cos t\\ \cos t& \sin t\end{array}\right)\left(\begin{array}{c}0\\ \cos (t)\cos (2t)\end{array}\right)=\left(\begin{array}{c}{\cos }^2(t)\cos (2t)\\ \sin (t)\cos (t)\cos (2t)\end{array}\right)$$
Note that ${\displaystyle v_2\left(t\right)=\sin \left(t\right)\cos \left(t\right)\cos \left(2t\right)=\frac{1}{2}\sin \left(2t\right)\cos \left(2t\right)=\frac{1}{4}\sin \left(4t\right),}$ so we can take
${\displaystyle U_2\left(t\right)=-\frac{1}{16}\cos \left(4t\right)}$
On other hand, we get
${\displaystyle u_1={\cos }^2\left(t\right)\cos \left(2t\right)=\cos \left(t\right)\frac{1}{2}(\cos \left(3t\right)+\cos \left(t\right))}$
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