UTILIZAR LA DEFINICION DE TRANSFORMADA DE LAPLACE Y RESOLVER LA SIGUIENTE FUNCION
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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](https://tex.z-dn.net/?f=%5Cdisplaystyle%0A%5Cmathcal%7BL%7D%5C%7BF%28t%29%5C%7D%28s%29%3D%5Cint_%7B0%7D%5E%7B%2B%5Cinfty%7De%5E%7B-st%7DF%28t%29%5C%2Cdt%5C%5C+%5C%5C+%5C%5C%0A%5Ctexttt%7BHagamos+los+c%5C%27alculos%3A%7D%5C%5C+%5C%5C%0A%5Cmathcal%7BL%7D%5C%7BF%28t%29%5C%7D%28s%29%3D%5Cint_%7B0%7D%5E%7B%2B%5Cinfty%7De%5E%7B-st%7D%5Cleft%282%5Csqrt%7B5%7Dt-5%5Ccosh%5Csqrt%7B5%7D%5C%2Ct%5Cright%29%5C%2Cdt%5C%5C+%5C%5C+%5C%5C%0A%5Cmathcal%7BL%7D%5C%7BF%28t%29%5C%7D%28s%29%3D2%5Csqrt%7B5%7D%5Cint_%7B0%7D%5E%7B%2B%5Cinfty%7Dt%5C%2Ce%5E%7B-st%7D%5C%2Cdt-5%5Cint_%7B0%7D%5E%7B%2B%5Cinfty%7De%5E%7B-st%7D%5C%2C%5Ccosh%5Csqrt%7B5%7D%5C%2Ct%5C%2Cdt%5C%5C+%5C%5C+%5C%5C%0A%5Cmathcal%7BL%7D%5C%7BF%28t%29%5C%7D%28s%29%3D2%5Csqrt%7B5%7D%5Cleft.%5Cleft%5B-%5Cdfrac%7Be%5E%7B-st%7D%7D%7Bs%5E2%7D-%5Cdfrac%7Bte%5E%7B-st%7D%7D%7Bs%7D%5Cright%5D%5Cright%7C_%7B0%7D%5E%7B%2B%5Cinfty%7D-5%5Cint_%7B0%7D%5E%7B%2B%5Cinfty%7De%5E%7B-st%7D%5C%2C%5Ccosh%5Csqrt%7B5%7D%5C%2Ct%5C%2Cdt%0A "\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")
![\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}}](https://tex.z-dn.net/?f=%5Cmathcal%7BL%7D%5C%7BF%28t%29%5C%7D%28s%29%3D%5Cdfrac%7B2%5Csqrt%7B5%7D%7D%7Bs%5E2%7D-%5Cdfrac%7B5%7D%7B2%7D%5Cleft.%5Cleft%5B-%5Cdfrac%7Be%5E%7B-%28s%2B%5Csqrt%7B5%7D%29t%7D%7D%7Bs%2B%5Csqrt%7B5%7D%7D%2B%5Cdfrac%7Be%5E%7B%28%5Csqrt%7B5%7D-s%29t%7D%7D%7B%5Csqrt%7B5%7D-s%7D%5Cright%5D%5Cright%7C_%7Bt%3D0%7D%5E%7Bt%5Cto%2B%5Cinfty%7D%5C%5C+%5C%5C+%5C%5C%0A%5Cmathcal%7BL%7D%5C%7BF%28t%29%5C%7D%28s%29%3D%5Cdfrac%7B2%5Csqrt%7B5%7D%7D%7Bs%5E2%7D-%5Cdfrac%7B5%7D%7B2%7D%5Cleft%5B%5Cdfrac%7B1%7D%7Bs%2B%5Csqrt%7B5%7D%7D%2B%5Cdfrac%7B1%7D%7Bs-%5Csqrt%7B5%7D%7D%5Cright%5D%5C%5C+%5C%5C+%5C%5C%0A%5Cboxed%7B%5Cmathcal%7BL%7D%5C%7BF%28t%29%5C%7D%28s%29%3D%5Cdfrac%7B2%5Csqrt%7B5%7D%7D%7Bs%5E2%7D-%5Cdfrac%7B5s%7D%7Bs%5E2-5%7D%7D%0A%0A "\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}}")
elenmora42:
muchisimas graciass!!
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