Question Number 94931 by mathmax by abdo last updated on 22/May/20
$$\mathrm{f}\:\mathrm{is}\:\mathrm{a}\:\mathrm{2}\left(×\right)\:\mathrm{derivable}\:\mathrm{function}\:\:\mathrm{and}\:\mathrm{L}\:\mathrm{laplace}\:\mathrm{transfom} \\ $$$$\mathrm{determine}\:\mathrm{L}\left(\mathrm{f}^{'} \right)\:\mathrm{a}\:\mathrm{L}\left(\mathrm{f}^{''} \right) \\ $$
Answered by mathmax by abdo last updated on 22/May/20
![L(f^′ (t))(p) =∫_0 ^∞ f^′ (t)e^(−pt) dt and by parts =[f(t)e^(−pt) ]_0 ^∞ −∫_0 ^∞ f(t)(−p)e^(−pt) dt =−f(0) +p ∫_0 ^∞ f(t)e^(−pt) dt =pL(f)−f(0^+ ) ⇒L(f^′ ) =pL(f)−f(0^+ ) also L(f^(′′) ) =L((f^′ )^′ ) =p L(f^′ )−f^′ (0^+ ) =p(pL(f)−f(0^+ ))−f^′ (0^+ ) =p^2 L(f)−pf(0^+ )−f^′ (0^+ )](https://www.tinkutara.com/question/Q95018.png)
$$\mathrm{L}\left(\mathrm{f}^{'} \left(\mathrm{t}\right)\right)\left(\mathrm{p}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\mathrm{f}^{'} \left(\mathrm{t}\right)\mathrm{e}^{−\mathrm{pt}} \:\mathrm{dt}\:\:\:\:\mathrm{and}\:\mathrm{by}\:\mathrm{parts}\: \\ $$$$=\left[\mathrm{f}\left(\mathrm{t}\right)\mathrm{e}^{−\mathrm{pt}} \right]_{\mathrm{0}} ^{\infty} \:−\int_{\mathrm{0}} ^{\infty} \:\mathrm{f}\left(\mathrm{t}\right)\left(−\mathrm{p}\right)\mathrm{e}^{−\mathrm{pt}} \:\mathrm{dt}\:=−\mathrm{f}\left(\mathrm{0}\right)\:+\mathrm{p}\:\int_{\mathrm{0}} ^{\infty} \:\mathrm{f}\left(\mathrm{t}\right)\mathrm{e}^{−\mathrm{pt}} \:\mathrm{dt} \\ $$$$=\mathrm{pL}\left(\mathrm{f}\right)−\mathrm{f}\left(\mathrm{0}^{+} \right)\:\Rightarrow\mathrm{L}\left(\mathrm{f}^{'} \right)\:=\mathrm{pL}\left(\mathrm{f}\right)−\mathrm{f}\left(\mathrm{0}^{+} \right)\:\mathrm{also} \\ $$$$\mathrm{L}\left(\mathrm{f}^{''} \right)\:=\mathrm{L}\left(\left(\mathrm{f}^{'} \right)^{'} \right)\:=\mathrm{p}\:\mathrm{L}\left(\mathrm{f}^{'} \right)−\mathrm{f}^{'} \left(\mathrm{0}^{+} \right) \\ $$$$=\mathrm{p}\left(\mathrm{pL}\left(\mathrm{f}\right)−\mathrm{f}\left(\mathrm{0}^{+} \right)\right)−\mathrm{f}^{'} \left(\mathrm{0}^{+} \right)\:=\mathrm{p}^{\mathrm{2}} \mathrm{L}\left(\mathrm{f}\right)−\mathrm{pf}\left(\mathrm{0}^{+} \right)−\mathrm{f}^{'} \left(\mathrm{0}^{+} \right) \\ $$