We first solve the characteristic polynomial: $r^3+r=0$ which yields $r_1=0, r_{2,3}=+- i$. So, solution to the general homogeneous equation is

$$y_g=u_1+u_2\cos(t)+u_3 \sin(t)$$

Idea in method of parameters is to let $u_1,u_2,u_3$ to be functions and then plot the $y_g$ into the equation (in this case compute it's first and third derivative), solving for $u_1,u_2,u_3$.

This can be done, but textbook page 242 gives you a nice final formula: $$u'_m=\frac{gW_m}{W}$$

So, we use this - maybe this was the reason why I finished 7 minutes earlier.

It turns out that $W(1,\cos(t), \sin(t))=1$. $W_m$ is just the same wronskian but the m-th column is replaced by the column $(0,0,1)$. With this definition

$$W_1=1, W_2=-\cos(t), W_3=-\sin(t)$$

Using the previously stated formula and the fact that in this case $g=\tan(t)$:

$$u_1'=\tan(t) \implies u_1=-\ln(\cos(t))$$

$$u_2'=-\tan(t)\cos(t)=-\sin(t)\implies u_2=\cos(t)$$

$$u_3'=-\tan(t)\sin(t)=-\left(\frac{1-\cos^2(t)}{\cos(t)} \right)=-\left(\frac{1}{\cos(t)}-\cos(t)\right)\implies

u_3=-\ln(\sec(t)+\tan(t))+\sin(t)$$

The third integral was the hardest. But in the homework problems you also had to compute $\int \frac{dt}{\cos(t)}$ which is surprisingly difficult. So, I had worked on the integral before and remembered the solution.

Now, we just have to plug $u_1,u_2,u_3$ back to the general solution of homogeneous equation $y_g$ and we obtain that the particular solution is:

$$Y_p=-\ln(\cos(t))+\cos^2(t)+\sin^2(t)-\sin(t)\ln(\sec(t)+tan(t))=-\ln(\cos(t))+1-\sin(t)\ln(\sec(t)+\tan(t))$$

General solution is therefore

$$Y_G=Y_p+c_1+c_2\cos(t)+c_3 \sin(t)$$