Find the solution of $$x''-x=f(t), \ \ t\geq 0, \ \ x(0)=1, \ \ x'(0)=0,$$ where $$f(t)=\begin{cases} 0 & 0\leq t\leq 1 \\ t-1 & t>1. \\ \end{cases} $$
First, I note that $f(t)$ can be re-written in terms of Heaviside functions: $f(t)=(t-1)H(t-1).$
Taking the Laplace transform of both sides: \begin{align} \mathcal{L}(x'')-\mathcal{L}(x)&=\mathcal{L}(f(t)) \\ s^2\mathcal{L}(x)-s-\mathcal{L}(x)&=\mathcal{L}((t-1)H(t-1)) \\ \mathcal{L}(x)&=\frac{\mathcal{L}((t-1)H(t-1))}{s^2-1}+\frac{s}{s^2-1}. \end{align} Using the Shifting Theorem, $$\mathcal{L}((t-1)H(t-1))=e^{-s}F(s)=\frac{e^{-s}}{s^2}.$$ Hence, \begin{align} \mathcal{L}(x)&=\frac{e^{-s}}{s^2(s^2-1)}+\frac{s}{s^2-1} \\ x(t)&=\mathcal{L^{-1}}\left(\frac{e^{-s}}{s^2(s^2-1)}+\frac{s}{s^2-1} \right)\\ &=\cosh(t)+\mathcal{L^{-1}}\left(\frac{e^{-s}}{s^2(s^2-1)}\right). \end{align} The answer given is $x(t)=\cosh(t)+H(t-1)(1-t+\sinh(t-1))$. I do not know how the last term was determined.