I'm asked to solve for $y(t)$ where
$$ y''(t) -y(t) = e^{-|t|} sgn(t) $$
where $-\infty < t < \infty$.
I thought I should Laplace transform both sides and try to get $Y(s) = … $ and inverse Laplace transform it back to get the answer; but I'm stuck on trying to Laplace transform $sgn(t)$. A given table of formulas doesn't contain it neither, it does however appear in Fourier transform, but then I'm unsure if I'm supposed to use Fourier transform to solve this problem instead?
Best Answer
The (unilateral) Laplace transform is defined by
$$\mathcal{L}\{f\}(s) = \int_{0}^{\infty} f(x)e^{-sx} \, dx. $$
Since this transform only concerns the interval $(0, \infty)$ it will only give a solution on this interval.
If you want to solve this problem by transform, the Fourier transform seems better suited for this problem. Since $y$ is integrable, taking Fourier transform lets you solve the equation in the space of tempered distributions. Then all you need to check that the solution obtained so is actually in $L^1$.
Other approach is to solve two equations
$$ \begin{cases} y_+''(t) - y_+(t) = e^{-t} & \text{on } (0, \infty) \\ y_+(0) = a, \\ y_+'(0) = b \end{cases} \qquad \text{and} \qquad \begin{cases} y_-''(t) - y_-(t) = -e^{t} & \text{on } (-\infty,0) \\ y_-(0) = a, \\ y_-'(0) = b \end{cases} $$
and concatenate them to create a global solution
$$ y(t) = \begin{cases} y_+(t) & t \geq 0 \\ y_-(t), & t \leq 0 \end{cases} $$
Not all choices of $(a, b)$ will make $y$ integrable, and indeed you can check that only the choice $(a, b) = (0, -\frac{1}{2})$ gives an integrable solution:
$$ y(t) = -\frac{t}{2} e^{-|t|}. $$