Tough Differential equation Can anyone help me solve this question ? 
$$ \large{y^{\prime \prime} + y = \tan{t} + e^{3t} -1}$$
I have gotten to a part when I know $r = \pm 1$ and then plugging them into a simple differential equation. I do not know how to the next step.
Thank you very much for all your help.

Find the complementary solution by solving
\begin{equation*}
y''+y=0.
\end{equation*}
Substitute $y=e^{\lambda t}$ to get 
\begin{equation*}
(\lambda ^2+1)e^{\lambda t}=0.
\end{equation*}
Therefore the zeros are $\lambda=i$ or $\lambda =-i.$ The general solution is given by 
\begin{equation*}
y=y_1+y_2=c_1e^{it}+\frac{c_2}{e^{it}}.
\end{equation*}
Apply Euler's identity and regroup the terms to get 
\begin{equation*}
y=(c_1+c_2)\cos(t)+i(c_1-c_2)\sin(t) \\
=c_3\cos(t)+c_2\sin(t).
\end{equation*}
For the particular solution, try $y_{b_1}=\cos(t)$ and $y_{b_2}=\sin(t).$ Calculating the Wronskian $W$ gives $1$. Let $f(t)$ be RHS of the differential equation. Use the two formulae
\begin{equation*}
v_1=-\int \frac{f(t)y_{b_2}}{W},~v_2=\int \frac{f(t)y_{b_1}}{W}
\end{equation*}
to get the particular solution
\begin{equation*}
y_p=v_1y_{b_1}+v_2y_{b_2}.
\end{equation*}
