Question

UTILIZAR LA DEFINICION DE TRANSFORMADA DE LAPLACE Y RESOLVER LA SIGUIENTE FUNCION

Answer 1

Bueno recordemos una parte de la definición ya que la función F es continua en toda la recta real.
 
        $\displaystyle \mathcal{L}\{F(t)\}(s)=\int_{0}^{+\infty}e^{-st}F(t)\,dt\\ \\ \\ \texttt{Hagamos los c\'alculos:}\\ \\ \mathcal{L}\{F(t)\}(s)=\int_{0}^{+\infty}e^{-st}\left(2\sqrt{5}t-5\cosh\sqrt{5}\,t\right)\,dt\\ \\ \\ \mathcal{L}\{F(t)\}(s)=2\sqrt{5}\int_{0}^{+\infty}t\,e^{-st}\,dt-5\int_{0}^{+\infty}e^{-st}\,\cosh\sqrt{5}\,t\,dt\\ \\ \\ \mathcal{L}\{F(t)\}(s)=2\sqrt{5}\left.\left[-\dfrac{e^{-st}}{s^2}-\dfrac{te^{-st}}{s}\right]\right|_{0}^{+\infty}-5\int_{0}^{+\infty}e^{-st}\,\cosh\sqrt{5}\,t\,dt$

       $\displaystyle \mathcal{L}\{F(t)\}(s)=\dfrac{2\sqrt{5}}{s^2}-5\int_{0}^{+\infty}e^{-st}\,\cosh\sqrt{5}\,t\,dt\\ \\ \\ \mathcal{L}\{F(t)\}(s)=\dfrac{2\sqrt{5}}{s^2}-5\int_{0}^{+\infty}e^{-st}\cdot\dfrac{e^{-\sqrt{5}t}+e^{\sqrt{5}t}}{2}\,dt\\ \\ \\ \mathcal{L}\{F(t)\}(s)=\dfrac{2\sqrt{5}}{s^2}-\dfrac{5}{2}\int_{0}^{+\infty}e^{-(s+\sqrt{5})t}+e^{(\sqrt{5}-s)t}\,dt$
 
        $\mathcal{L}\{F(t)\}(s)=\dfrac{2\sqrt{5}}{s^2}-\dfrac{5}{2}\left.\left[-\dfrac{e^{-(s+\sqrt{5})t}}{s+\sqrt{5}}+\dfrac{e^{(\sqrt{5}-s)t}}{\sqrt{5}-s}\right]\right|_{t=0}^{t\to+\infty}\\ \\ \\ \mathcal{L}\{F(t)\}(s)=\dfrac{2\sqrt{5}}{s^2}-\dfrac{5}{2}\left[\dfrac{1}{s+\sqrt{5}}+\dfrac{1}{s-\sqrt{5}}\right]\\ \\ \\ \boxed{\mathcal{L}\{F(t)\}(s)=\dfrac{2\sqrt{5}}{s^2}-\dfrac{5s}{s^2-5}}$