Solve the following IVP.
$$y''(t)+y'(t)+\frac{5}{4}y(t)=g(t) \text{ where }g(t)= \begin{cases} \sin(t) & 0\leq t<\pi \\ 0 & t\geq\pi \end{cases}\text{ and }y(0)=y'(0)=0 $$
My procedure:
For $t\geq0:$
$$ y''(t)+y'(t)+\frac{5}{4}y(t)=\sin(t)+\sin(t-\pi)\theta(x-\pi) \\ \mathcal{L}\left\{y''(t)+y'(t)+\frac{5}{4}y(t)\right\}= \mathcal{L}\left\{\sin(t)+\sin(t-\pi)\theta(x-\pi)\right\} \\ \left(s^2+s+\frac54\right)\mathcal{L}\left\{y(t)\right\}=\frac1{s^2+1}+\frac{e^{-\pi s}}{s^2+1}\\ \xrightarrow{\text{partial fractions}} \\ y(t)=\mathcal{L}^{-1}\left\{-\frac{16}{17}\left(\frac{s}{s^2+1}\right)+\frac{4}{17}\left(\frac{1}{s^2+1}\right)+\frac{16}{17}\left(\frac{s+\frac12-\frac12}{\left(s+\frac12\right)^2+1}\right)+ \frac{12}{17}\left(\frac{s}{\left(s+\frac12\right)^2+1}\right)+e^{-\pi s}\left(-\frac{16}{17}\left(\frac{s}{s^2+1}\right)+\frac{4}{17}\left(\frac{1}{s^2+1}\right)+\frac{16}{17}\left(\frac{s+\frac12-\frac12}{\left(s+\frac12\right)^2+1}\right)+\frac{12}{17}\left(\frac{s}{\left(s+\frac12\right)^2+1}\right)\right) \right\} $$ After taking the inverse Laplace Transform of the above expression, we obtain the following (I will not write the calculations since they will occupy too much space): $$\boxed{y(t)=\frac{4}{17}\left(e^{-t/2}-e^{(\pi-t)/2}\theta(t-\pi)-\theta(t-\pi)+1\right)\sin(t)++\frac{16}{17}\left(e^{-t/2}-e^{(\pi-t)/2}\theta(t-\pi)+\theta(t-\pi)-1\right)\cos(t)}$$ Would this be correct? I cannot check my work so it is difficult for me to find out if I am right.
I think you mean to write
$$ y(t) = \frac{4}{17}\left(e^{-t/2}-e^{-(t-\pi)/2}\theta(t-\pi)-\theta(t-\pi)+1\right)\sin t + \\ \frac{16}{17}\left(e^{-t/2}-e^{-(t-\pi)/2}\theta(t-\pi)+\theta(t-\pi)-1\right)\cos t $$
this can be rewritten as
$$ y(t) = \begin{cases} \displaystyle \frac{4}{17}(e^{-t/2}+1)\sin t + \frac{16}{17}(e^{-t/2}-1)\cos t , && 0 < t < \pi \\ \displaystyle -(e^{\pi/2}-1)e^{-t/2}\left( \frac{4}{17}\sin t + \frac{16}{17}\cos t \right), && t > \pi \end{cases} $$
By inspection you have a particular solution $$ y_p(t) = -\frac{16}{17}\cos t + \frac{4}{17}\sin t $$
that satsifies $$ y'' + y + \frac{5}{4}y = \sin t $$
and the homogeneous solution of the form $$ y_h(t) = e^{-t/2}(a\cos t + b\sin t) $$
for $$ y'' + y + \frac{5}{4}y = 0 $$
Also, the first derivative is $$ y'(t) = \begin{cases} \displaystyle \left(-\frac{4}{17}e^{-t/2} + \frac{4}{17} \right)\cos t - \left(\frac{18}{17}e^{-t/2}-\frac{16}{17} \right)\sin t, && 0 < t < \pi \\ \displaystyle (e^{\pi/2}-1)e^{-t/2}\left(\frac{4}{17}\cos t + \frac{18}{17}\sin t \right), && t > \pi \end{cases} $$
The initial condition $y(0) = y'(0) = 0$ and continuity for $y(\pi)$, $y'(\pi)$ are all satisfied