Solve $\frac{d^2u}{dt^2} + \omega^2 u = \sin(\lambda t)$ by using the following algorithm... Solve $$\frac{d^2u}{dt^2} + \omega^2 u = \sin(\lambda t)$$
By using the following integral to create the particular solution $Y_P$ like this...
$$Y_P = \int_0^t G(t-s) f(s) ds$$ where $$G(t) = \int_0^t e^{\alpha (t-s)} e^{\beta s} ds$$
where $\alpha$ and $\beta$ are the roots of the auxillary equation of the homogenous form of the D.E. above.
I'm totally stuck because I only know how to solve by using undetermined coeff. or variation of parameters methods. 
What is this solving algorithm and how to solve by using it????
 A: Substitute the proposed solution in the equation as $u$. But before that note that by evaluating the integral:
\begin{equation*}
 G(t) = \frac{e^{\beta t} - e^{\alpha t}}{\beta - \alpha}, \hspace{0.5cm} where \hspace{0.5cm} \beta = j\omega, \alpha = -j\omega
\end{equation*}
Now, if we substitute $u(t) = \int_{0}^{t}G(t-s)f(s)ds$, we need first to find second time derivative of $u$. By leibniz rule:
\begin{equation*}
 \frac{du}{dt} = G(0)f(t) + \int_{0}^{t}\frac{\partial G(t-s)f(s)}{\partial t}ds = \int_{0}^{t} \frac{\beta e^{\beta (t-s)} - \alpha e^{\alpha (t-s)}}{\beta - \alpha}f(s)ds 
\end{equation*}
\begin{equation*}
\frac{d^2 u}{dt^2} = \frac{d}{dt}(\int_{0}^{t} \frac{\beta e^{\beta (t-s)} - \alpha e^{\alpha (t-s)}}{\beta - \alpha}f(s)ds ) = f(t) + \int_{0}^{t} \frac{\beta^2 e^{\beta (t-s)} - \alpha^2 e^{\alpha (t-s)}}{\beta - \alpha}f(s)ds 
\end{equation*}
Now substituting $\frac{d^2 u}{dt^2}$ in your differential equation, and noting that $\beta = j\omega, \alpha = -j\omega$, we have:
\begin{equation*}
\frac{d^2u}{dt^2} + \omega ^2u(t) = f(t) + \int_{0}^{t} \frac{\beta^2 e^{\beta (t-s)} - \alpha^2 e^{\alpha (t-s)}}{\beta - \alpha}f(s)ds + \omega ^2 \int_{0}^{t} \frac{e^{\beta (t-s)} - e^{\alpha (t-s)}}{\beta - \alpha}f(s)ds = f(t) + \int_{0}^{t} \frac{(\beta^2 + \omega ^2) e^{\beta (t-s)} - (\alpha^2  + \omega^2) e^{\alpha (t-s)}}{\beta - \alpha}f(s)ds = sin(\lambda t)
\end{equation*}
Here, as $(\beta^2 + \omega ^2) = (\alpha^2 + \omega ^2) =0$, we have:
\begin{equation*}
f(t) = sin(\lambda t).
\end{equation*}
Then,  your $Y_p$ is:
\begin{equation*}
Y_p = \int_{0}^{t}  \frac{e^{\beta (t-s)} - e^{\alpha (t-s)}}{\beta - \alpha} sin(\lambda s)ds 
\end{equation*}
