Differential Equations with Discontinuous Forcing Functions $$ y''+y'+1.25y = g(t), \quad t > 0, $$
$$y(0) = 0, \quad  y'(0) = 0 $$
$$g(t) = \left\{ \begin{array}{ll}
\sin{t}  &  0 \le t < \pi  \\ 
0        &  t \ge \pi
\end{array}\right.$$
Solve the initial value problem. Draw the graphs of the solution and of the forcing function; explain how they are related. 
I am confused about what a forcing function is and how to solve it. I attempted to solve it piece wise, finding that the solution for $y'' + y' + 1.25y = \sin{t}$  is $\frac{4}{17} \sin{t} - \frac{16}{17} \cos{t}$. I got the general solution for $y'' + y' + 1.25y = 0$ as $y = ce^{-1/2t}cos(t) + ce^{-1/2t} \sin{t}$. I wasn't sure what to do next.
Was I supposed to use laplace transform instead? 
 A: Your equation is a 2nd order, constant coefficients, non-homogenous and linear ODE. Furthermore, the non-homogenous term is a piecewise function that, since $t > 0$ splits your domain into two subdomains, $I_1 = (0,\pi)$ and $I_2 = [\pi,\infty)$. This leads you to solve the equation for each of the subdomain. For doing that, follow, for example, this link or this. Notice, that you will end up with two solutions, $y_1(x)$ for $x \in I_1$ and $y_2$ for $x \in I_2$ with, in addition, four (different) constants of integration. You can then put some of them as functions of the others in order to have a continous solution, provided the intial conditions in $x\in I_1$. 
Notice also that the homogenous part of your equations doesn't change, so it should remain the same. Indeed, we have:
$$\mathrm{L}[y_h] = y'' +y' + \frac{5}{4} y = 0, $$ and the characteristic equation tells us that $r_{1,2} = - \frac{1}{2} \pm \mathrm{i}$ are its respective solutions. So:
$$y_h(t) = e^{-t/2} ( A\cos{t} + B \sin{t} ).$$

Let's see how we can deal with this problem using Laplace transform. Taking Laplace transform on both sides we arrive at:
$$Y(s)(s^2 + s+ 5/4) = \mathcal{L}_s g(t), $$ where intial conditions have been considered and the RHS is the Laplace transform of a function defined over two intervals, i.e.:
$$\mathcal{L}_s g(t) = \int^\infty_0 g(t)e^{-s t} \, \mathrm{d}t = \int^\pi_0 \sin{t}e^{-st} \, \mathrm{d}t+\int^\infty_\pi 0 \times e^{-st} \, \mathrm{d}t = \frac{1+e^{-\pi s}}{s^2+1},$$ 
thus, solving for $Y(s)$ and anti-Laplace-transforming both sides we get:

$$\color{blue}{y(t) = \mathcal{L}^{-1}[Y(s)] = \mathcal{L}^{-1} \left[ \frac{1 + e^{-\pi s}}{(s^2+s+5/4)(s^2+1) }   \right] = \mathcal{L}^{-1}[G(s)].}$$

Note now that we can write $G(s)$ as follows:
$$ G(s) = \frac{1}{ (s^2+s+5/4)(s^2+1)   } + \frac{e^{-\pi s}}{(s^2+s+5/4)(s^2+1)},  $$
and, since $1/(s^2+1) = \mathcal{L}_s[\sin{t}]$ and $1/(s^2+s+5/4) = \mathcal{L}_s[e^{-t/2} \sin{t} ]$ we can write:
$$ G(s) = \underbrace{\mathcal{L}_s[\sin{t}]\mathcal{L}_s[e^{-t/2} \sin{t} ]}_{Q(s)}  \ +  \ e^{-\pi s} \mathcal{L}_s[\sin{t}]\mathcal{L}_s[e^{-t/2} \sin{t} ]. $$
Therefore, by the convolution theorem we have:
\begin{align}
 q(t)  & = \sin{t} * e^{-t/2} \sin{t} = \int^t_0 \sin{(t-\tau)} e^{ -t/2} \sin{t} \, \mathrm{d} \tau \\
  & =  \frac{8}{17} e^{-t/4} \left(\cosh(t/4) \sin{t} - 4 \cos{t} \sinh{(t/4)} \right),  
\end{align} 
where Mathematica has made things easier here. Finally, using the fact that $\mathcal{L}^{-1} [F(s) e^{-a s}]=H(t-a) f(t-a), $ $a\in \mathbb{R}$, where $H$ is the Heaviside or unit step function, we finally obtain the solution as follows:

$$\color{blue}{y(t) = q(t) + H(t-\pi) q(t-\pi)}.$$

Here is a sketch of the solution for $0 \leq t \leq 5 \pi$ obtained via Laplace transform which matches, of course, with that obtained using $\texttt{DSolve}$ with Mathematica:

we can see that, if this corresponds to a dynamical system, then it is a stable damped harmonic oscillator. 

Cheers!
A: You already proved that the solution on $(0,\pi)$ is $$y(t)=\frac{4}{17} \sin{t} - \frac{16}{17} \cos{t}+A\mathrm e^{-t/2}\cos t + B\mathrm e^{-t/2} \sin{t},$$ for some constants $A$ and $B$. Your next step should be to determine $(A,B)$ using the initial conditions $y(0)$ and $y'(0)$.
You (nearly) already proved that the solution on $(\pi,\infty)$ is $$y(t)=C\mathrm e^{-t/2}\cos t + D\mathrm e^{-t/2} \sin{t},$$ for some constants $C$ and $D$. Your next step should be to determine $(C,D)$ using the continuity of the functions $y$ and $y'$ at $\pi$. In other words, one should have $$\lim\limits_{t\to\pi,t\lt\pi}y(t)=\lim\limits_{t\to\pi,t\gt\pi}y(t),\qquad\lim\limits_{t\to\pi,t\lt\pi}y'(t)=\lim\limits_{t\to\pi,t÷gt\pi}y'(t).$$
