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Question Number 95838 by mathmax by abdo last updated on 28/May/20
$$\mathrm{determine}\:\mathrm{L}\left(\mathrm{f}^{\left(\mathrm{3}\right)} \left(\mathrm{x}\right)\:\:\mathrm{with}\:\mathrm{L}\:\mathrm{is}\:\mathrm{laplace}\:\mathrm{transform}\right. \\ $$
Answered by mathmax by abdo last updated on 28/May/20
$$\mathrm{L}\left(\mathrm{f}^{\left(\mathrm{3}\right)} \left(\mathrm{x}\right)\right)\:=\mathrm{L}\left(\left(\mathrm{f}^{'} \right)^{''} \right)\:=\mathrm{x}^{\mathrm{2}} \mathrm{L}\left(\mathrm{f}^{'} \right)−\mathrm{x}\:\mathrm{f}^{'} \left(\mathrm{0}\right)−\mathrm{f}^{''} \left(\mathrm{0}\right) \\ $$$$=\mathrm{x}^{\mathrm{2}} \left(\:\mathrm{xL}\left(\mathrm{f}\right)−\mathrm{f}\left(\mathrm{0}\right)\right)−\mathrm{xf}^{'} \left(\mathrm{0}\right)−\mathrm{f}^{''} \left(\mathrm{0}\right) \\ $$$$=\mathrm{x}^{\mathrm{3}} \:\mathrm{L}\left(\mathrm{f}\right)−\mathrm{x}^{\mathrm{2}} \mathrm{f}\left(\mathrm{0}\right)−\mathrm{xf}^{'} \left(\mathrm{0}\right)−\mathrm{f}^{''} \left(\mathrm{0}\right) \\ $$