Question Number 184307 by Mastermind last updated on 05/Jan/23
$$\mathrm{Show}\:\mathrm{that}\:\mathrm{the}\:\mathrm{boundary}−\mathrm{value} \\ $$$$\mathrm{problem}\:\mathrm{y}''+\lambda\mathrm{y}=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{0}, \\ $$$$\mathrm{y}\left(\mathrm{L}\right)=\mathrm{0}\:\mathrm{has}\:\mathrm{only}\:\mathrm{the}\:\mathrm{trival}\:\mathrm{solution} \\ $$$$\mathrm{y}=\mathrm{0}\:\mathrm{for}\:\mathrm{the}\:\mathrm{cases}\:\lambda=\mathrm{0}\:\mathrm{and}\:\lambda<\mathrm{0}. \\ $$$$\mathrm{let}\:\mathrm{L}\:\mathrm{be}\:\mathrm{a}\:\mathrm{non}−\mathrm{zero}\:\mathrm{real}\:\mathrm{number}. \\ $$$$ \\ $$$$ \\ $$$$? \\ $$
Answered by FelipeLz last updated on 05/Jan/23
$${y}''+\lambda{y}\:=\:\mathrm{0} \\ $$$${y}\:=\:{e}^{{r}} \:\rightarrow\:{r}^{\mathrm{2}} {e}^{{r}} +\lambda{e}^{{r}} \:=\:\mathrm{0} \\ $$$${r}^{\mathrm{2}} \:=\:−\lambda \\ $$$$\: \\ $$$$\lambda\:=\:\mathrm{0}\:\rightarrow\:{r}\:=\:\mathrm{0}\:\therefore\:{y}\left({x}\right)\:=\:{c}\: \\ $$$${y}\left(\mathrm{0}\right)\:=\:\mathrm{0}\:=\:{c} \\ $$$${y}\left({x}\right)\:=\:\mathrm{0} \\ $$$$\: \\ $$$$\lambda\:<\:\mathrm{0}\:\rightarrow\:{r}\:=\:\pm\sqrt{−\lambda}\:\therefore\:{y}\left({x}\right)\:=\:{c}_{\mathrm{1}} {e}^{\sqrt{−\lambda}{x}} +{c}_{\mathrm{2}} {e}^{−\sqrt{−\lambda}{x}} \\ $$$${y}\left(\mathrm{0}\right)\:=\:\mathrm{0}\:\rightarrow\:{c}_{\mathrm{1}} +{c}_{\mathrm{2}} \:=\:\mathrm{0}\:\therefore\:{y}\left({x}\right)\:=\:{c}\left({e}^{\sqrt{−\lambda}{x}} −{e}^{−\sqrt{−\lambda}{x}} \right) \\ $$$${y}\left({L}\right)\:=\:\mathrm{0}\:\rightarrow\:{c}\left({e}^{\sqrt{−\lambda}{L}} −{e}^{−\sqrt{−\lambda}{L}} \right)\:=\:\mathrm{0}\: \\ $$$${L}\:\neq\:\mathrm{0}\:\rightarrow\:{e}^{\sqrt{−\lambda}{L}} −{e}^{−\sqrt{−\lambda}{L}} \:\neq\:\mathrm{0}\:\therefore\:{c}\:=\:\mathrm{0} \\ $$$${y}\left({x}\right)\:=\:\mathrm{0} \\ $$$$ \\ $$