Laplace transform of the cos hyperbolic function

Posted December 8th, 2007 by Structure

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

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

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

Comments

On May 18th, 2012 emmar says:

This blog has got lots of really helpful information on it! Cheers for sharing it with me! Luxury projects in gurgaon