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Question Number 184307 by Mastermind last updated on 05/Jan/23

$$\mathrm{Show}\mathrm{that}\mathrm{the}\mathrm{boundary}-\mathrm{value}$$ $$\mathrm{problem}{\mathrm{y}}^{″}+\lambda \mathrm{y}=0\mathrm{y}\left(0\right)=0,$$ $$\mathrm{y}\left(\mathrm{L}\right)=0\mathrm{has}\mathrm{only}\mathrm{the}\mathrm{trival}\mathrm{solution}$$ $$\mathrm{y}=0\mathrm{for}\mathrm{the}\mathrm{cases}\lambda =0\mathrm{and}\lambda <0.$$ $$\mathrm{let}\mathrm{L}\mathrm{be}\mathrm{a}\mathrm{non}-\mathrm{zero}\mathrm{real}\mathrm{number}.$$ $$$$ $$$$ $$?$$

Answered by FelipeLz last updated on 05/Jan/23

$$y^{″}+\lambda y=0$$ $$y=e^r\to r^2{e}^r+\lambda e^r=0$$ $$r^2=-\lambda$$ $$$$ $$\lambda =0\to r=0\therefore y\left(x\right)=c$$ $$y\left(0\right)=0=c$$ $$y\left(x\right)=0$$ $$$$ $$\lambda <0\to r=\pm \sqrt{-\lambda }\therefore y\left(x\right)=c_1{e}^{\sqrt{-\lambda }x}+c_2{e}^{-\sqrt{-\lambda }x}$$ $$y\left(0\right)=0\to c_1+c_2=0\therefore y\left(x\right)=c(e^{\sqrt{-\lambda }x}-e^{-\sqrt{-\lambda }x})$$ $$y\left(L\right)=0\to c(e^{\sqrt{-\lambda }L}-e^{-\sqrt{-\lambda }L})=0$$ $$L\ne 0\to e^{\sqrt{-\lambda }L}-e^{-\sqrt{-\lambda }L}\ne 0\therefore c=0$$ $$y\left(x\right)=0$$ $$$$

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