Laplace transform of the cos hyperbolic function

Function $ f:R\rightarrow R $ , $ f(t)=\cosh t\:h(t) $ verifies differential equation

$$x"(t)-x(t)=0\;\:x(0+)=1,\:\:x'(0+)=0$$
$$\mathcal{L}(x(t)(s)=X(s)$$
$$\mathcal{L}(x'(t)(s)=sX(s)-1$$
$$\mathcal{L}(x"\;(t)(s)=s^2X(s)-s$$

We have

$$s^2X(s)-s-X(s)=0$$

so

$$X(s)=\mathcal{L}(\cosh t\:h(t))(s)=\frac{s}{s^2-1}$$
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