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Question Number 94931 by mathmax by abdo last updated on 22/May/20

$$\mathrm{f}\mathrm{is}\mathrm{a}2\left(\times \right)\mathrm{derivable}\mathrm{function}\mathrm{and}\mathrm{L}\mathrm{laplace}\mathrm{transfom}$$$$\mathrm{determine}\mathrm{L}\left({\mathrm{f}}^{{}^{\prime }}\right)\mathrm{a}\mathrm{L}\left({\mathrm{f}}^{{}^{″}}\right)$$

Answered by mathmax by abdo last updated on 22/May/20

$$\mathrm{L}\left({\mathrm{f}}^{{}^{\prime }}\left(\mathrm{t}\right)\right)\left(\mathrm{p}\right)={\int }_0^{\mathrm{\infty }}{\mathrm{f}}^{{}^{\prime }}\left(\mathrm{t}\right){\mathrm{e}}^{-\mathrm{pt}}\mathrm{dt}\mathrm{and}\mathrm{by}\mathrm{parts}$$$$={\left[\mathrm{f}\left(\mathrm{t}\right){\mathrm{e}}^{-\mathrm{pt}}\right]}_0^{\mathrm{\infty }}-{\int }_0^{\mathrm{\infty }}\mathrm{f}\left(\mathrm{t}\right)(-\mathrm{p}){\mathrm{e}}^{-\mathrm{pt}}\mathrm{dt}=-\mathrm{f}\left(0\right)+\mathrm{p}{\int }_0^{\mathrm{\infty }}\mathrm{f}\left(\mathrm{t}\right){\mathrm{e}}^{-\mathrm{pt}}\mathrm{dt}$$$$=\mathrm{pL}\left(\mathrm{f}\right)-\mathrm{f}\left(0^{+}\right)\Rightarrow \mathrm{L}\left({\mathrm{f}}^{{}^{\prime }}\right)=\mathrm{pL}\left(\mathrm{f}\right)-\mathrm{f}\left(0^{+}\right)\mathrm{also}$$$$\mathrm{L}\left({\mathrm{f}}^{{}^{″}}\right)=\mathrm{L}\left({\left({\mathrm{f}}^{{}^{\prime }}\right)}^{{}^{\prime }}\right)=\mathrm{p}\mathrm{L}\left({\mathrm{f}}^{{}^{\prime }}\right)-{\mathrm{f}}^{{}^{\prime }}\left(0^{+}\right)$$$$=\mathrm{p}(\mathrm{pL}\left(\mathrm{f}\right)-\mathrm{f}\left(0^{+}\right))-{\mathrm{f}}^{{}^{\prime }}\left(0^{+}\right)={\mathrm{p}}^2\mathrm{L}\left(\mathrm{f}\right)-\mathrm{pf}\left(0^{+}\right)-{\mathrm{f}}^{{}^{\prime }}\left(0^{+}\right)$$

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