Piecewise Forcing Function Differential With Initial Values at Negative Infinity I have the following problem:
$$x''+\omega ^2 x=F(t), \,\, x(-\infty)=x'(-\infty)=0$$
$$F(t)=\begin{cases} 1 \text  { for } t\geq 0 \\ 0 \text  { for } t < 0  \end{cases}$$
And I am confused on the piece-wise aspect of it, as well as it's initial values at negative infinity. In my calc 4 class, to solve a function like this, we would have used a Laplace. This seems (to me) to fail because our initial conditions are not at 0.
What I've tried:
The homogenous solution:
$c_1sin(wt) + c_2cos(wt)$ or $c_1e^{iwt} + c_2e^{-iwt}$.
I don't know how to find $c_1$ and $c_2$ Is our solution going to be piecewise? I would assume that's not optimal. I know on the right (t>=0), that $1/w^2$ is a solution. 
I spoke to my professor a bit and he mentioned two things- One, that we could just use $t_0$ in place of infinity, and get a solution in terms of $t_0$, and later plug in, and also that we would want to make the left side and the right side continuous at the point they meet. I just don't know what to do.
 A: Since the equation is piece-wise, it's safe to assume the solution will be piece-wise as well.
For $t < 0$, you have the IVP
$$ x'' + \omega^2 x = 0, \quad x(-\infty) = x'(-\infty) = 0 $$
The only solution that satisfies the initial values is the zero function $x(t < 0) = 0$. Maintaining continuity and smoothness, we obtain the IVP for $t \ge 0$
$$ x'' + \omega^2 x = 1, \quad x(0) = x'(0) = 0 $$
This is easily solvable using undetermined coefficients, with the general solution being
$$ x(t\ge 0) = \frac{1}{\omega^2} + c_1\cos(\omega x) + c_2\sin(\omega x) $$
The final solution is
$$ x(t) = \begin{cases} 0, && t < 0 \\ \dfrac{1 - \cos (\omega x)}{\omega^2}, && t \ge 0 \end{cases} $$
Note that if $\omega=0$, we must take the limit of the above expression as $\omega \to 0$
$$ x(t) = \begin{cases} 0, && t < 0 \\ \dfrac{x^2}{2}, && t \ge 0 \end{cases} $$

It is also possible to solve using the Laplace transform with your professor's hint. We first look for a solution of the following IVP
$$ y'' + \omega^2 y = u(t-t_0), \quad y(0) = y'(0) = 0 $$
where $u(t)$ is the Heaviside step function, such that
$$ u(t-t_0) = \begin{cases} 0, && t < t_0 \\ 1, && t \ge t_0 \end{cases} $$
We obtain the Laplace transform of the solution
$$ Y(s) = \frac{e^{-t_0s}}{s(s^2+\omega^2)} = \frac{e^{-t_0s}}{\omega^2}\left( \frac{1}{s} - \frac{s}{s^2+\omega^2}\right) $$
Inverse transforming back
$$ y(t) = \frac{1}{\omega^2} u(t-t_0) \big[1 - \cos\big(\omega(t-t_0)\big)\big] $$
Shifting the initial condition from $0$ to $-t_0$ results in the original equation. Therefore
$$ x(t) = y(t+t_0) = \frac{1}{\omega^2} u(t) \big[1 - \cos (\omega t)\big] $$
We can take the limit $t_0 \to \infty$, but this turns out to be irrelevant.
